202 + Interview Questions in SYSTEMS-OF-PARTICLES-AND-ROTATIONAL-MOTION IN GENERAL AWARENESS Page 1 InterviewSolution

1.	What is the position vector of centre of mass of two particles of equal masses ?
Answer» It is the averge of the position vectors to two particles.

Discussion

2.	Consider a two particle system with particles having masses `m_1 and m_2` if the first particle is pushed towards the centre of mass through a distance d, by what distance should the second particle is moved, so as to keep the center of mass at the same position?A. `d`B. `(m_(2)//m_(1))d`C. `[m_(1)//(m_(1) + m_(2))]d`D. `(m_(1)//m_(2))d`
Answer» Correct Answer - D Let the distance of particle of mass `m_(2)` be `d_(2)` and distance of particle of mass `m_(2)` be `d_(1) = d`. In order to keep mass centre at the original position `m_(1)d_(1) = m_(2)d_(2)` or `d_(2) = (m_(1)d_(1))/(m_(2)) = ((m_(1))/(m_(2)))d` as `d_(1) = d`

Discussion

3.	By convention, anticlockwise moments are ………and…………are taken as…………. .
Answer» taken as positive , clockwise moments , negative.

Discussion

4.	As shown in Fig. the two sides of a step ladder `BA` and `CA` are `1.6 m` long and hinged at `A`. A rope `DE, 0.5 m` is tied half way up. A weight `40 kg` is suspended from a point `F, 1.2 m` from B along the ladder` BA`. Assuming the floor to be fricionless and neglecting the weight of the ladder, find the tension in the rope and forces exerted by the floor on the ladder. (Take `g = 9.8 m//s^(2))` (Hint. Consider the eqilibrium of each side of the ladder separately.)
Answer» Here, `BA = CA = 1.6 m, DE = 0.5 m, M = 40 kg , BF = 1.2 m` Let `T =` tension in the rope, `N_(1),N_(2) =` normal reaction at `B and C` respectively, i.e. forces extered by the floor on the ladder. In Fig. we find, As `DE = 0.5m , BC = 10 m , FH = (1)/(2)DG = (1)/(2) xx 0.25 = 0.125 m` `AG = sqrt(AD^(2) - DG^(2)) = sqrt(0.8^(2) - 0.25^(2)) = 0.76 m` For translational equilibrium of the step ladder, `N_(1) + N_(2) - Mg = 0` or `N_(1) + N_(2) = Mg = 40 xx 9.8 = 392` ..(i) For rotational equilibrium of the step ladder, taking moments about `A`, we find `-N_(1) xx BK + Mg xx FH + N_(2) xx CK + T xx AG - T xx AG = 0` `-N_(1) xx 0.5 + 40 xx 9.8 xx 0.125 + N_(2) xx 0.5 = 0 or (N_(1) - N_(2)) 0.5 = 40 xx 9.8 xx 0.125` Add (i) and (ii), `2N_(1) = 490` ...(ii) `N_(1) = (490)/(2) = 245N` From (i), `N_(2) = 392 - N_(1) = 392 - 245 = 147 N` For rotational equilibrium of side `AB` of the step ladder, taking moments about `A`, we get `Mg xx FH - N_(1) xx BK + T xx AG = 0` `40 xx 9.8 xx 0.125 xx 0.5 + T xx 0.76 = 0` `:. T xx 0.76 = 245 xx 0.5 - 40 xx 9.8 xx 0.125 = 122.5 - 49 = 73.5` `T = (73.5)/(0.76) = 96.7N`

Discussion

5.	A uniform square plate and a disc having same mass per unit area are kept in contact as shown in Fig. The side of square and diameter of circle are both equal to `L`. Locate the position of centre of mass of the system w.r.t. the centre of the square.
Answer» Let mass per unit area of the square plate = mass per unit area of disc `= sigma` `:.` Mass of square plate `m_(1) = sigmaL^(2)`, and mass of disc `m_(2) = (pir^(2)) sigma = (pi(L^(2))/(4)) sigma` Centre of square is `O_(1)`, where `m_(1)` is concentrated and centre of disc is `O_(2)`, where `m_(2)` is concentrated. If `x` is distance of c.m of the combination from `O_(1)`, then `x_(cm) = (m_(1)x_(1) + m_(2)x_(2))/(m_(1) + m_(2)) = (m_(1) xx 0 + m_(2)L)/(m_(1) + m_(2))` `=((pi(L^(2))/(4))sigma xx L)/(sigmaL^(2) + (pi L^(2)sigma)/(4)) = (sigmapi L^(2)//4 xx L)/(sigmaL^(2)(1 + pi//4))` `x_(cm) = (pi//4 xx L)/((4 + pi)//4) = ((piL)/(4 + pi))`

Discussion

6.	A disc of radius `0.5 m` is rotating about an axis passing through its centre and perpendicular to its plane. A tangential force of `2000 N` is applied to bring the dics to rest `2 s`. Calculate its abgular momentum.
Answer» Here, `r = 0.5m, F = 2000N, t = 2 s` Final angular momentum, `L_(2) = 0`, Initial angular momentum, `L_(1) = ?` torque applied, `tau = - F xx r = - 2000 xx 0.5` `= - 1000 N-m` As `tau= (L_(2) - L_(1))/(t)` `:. - 1000 = (0 - L_(1))/(2), L_(1) = 2000 kg m^(2) s^(-1)`

Discussion

7.	A body of moment of inertia `0.5 kg m^(2)` is rotating about a given axis at the rate `1 rps`. What is kinetic energy of rotation of the body about that axis ?
Answer» Here, `I = 0.5kg m^(2), n = 1 rps, KE = ?` `KE` of rotation `= (1)/(2) I omega^(2) = (1)/(2)I (2pi n)^(2) = 2pi^(2) n^(2) I = 2 xx (22)/(7) xx (22)/(7) xx 1^(2)xx 0.5 = 9.85 J`

Discussion

8.	A cylinder of length `20 cm` and radius `10 cm` is rotating about its central axis at an angular speed of `100 rad//s`. What tangential force will stop the cylinder at a unifrom rate is `10 s` ? Given moment of inertia of the cylinder about its axis of rotation is `8.0 kg m^(2)`.
Answer» Here, `l= 20 cm, r = 10 cm, omega_(1) = 100 rad//s`, `F = ?, omega_(2) =0, t = 10 s, I = 0.8 kg m^(2)` `tau = F xx r = I alpha = (I(omega_(2) - omega_(1)))/(t)` `:. F = (I (omega_(2) - omega_(1)))/(t xx r) = (0.8 (0 - 100))/(10 xx 10//100) = - 80 N`

Discussion

9.	A dics rotates about the central axis starting from rest and accelerates with constant angular accelertion. At one time, it is rotating at `10 rps` , `60` revoluting later, its angular speed is `15rps`. Calculate (i) angular acceleration (ii) time required to complate `60` revols (iii) the time required to reach `10 rev//sec` angular speed and (iv) number of revoluting from rest until the time the disc reaches `10rps` angular speed.
Answer» Here, `omega_(0) = 10rps = 20 pi rad//s` `theta = 60 revols = 60 xx 2 pi radian` `omega = 15 rps = 15 xx 2 pi rad//s` From `omega^(2) - omega_(0)^(2) = 2 alpha theta` `alpha = (omega^(2) - omega_(0)^(2))/(2 theta) = ((30 pi)^(2) - (20 pi)^(2))/(2 xx 60 xx 2pi)` `alpha = (500 pi^(2))/(240 pi) = (25)/(12) xx 3.14 rad//s^(2)` `= 6.54 rad//s^(2)` From `omega = omega_(0) + alpha t` `t = (omega - omega_(0))/(alpha) = (30pi - 20 pi)/(6.54)` `t = (10 xx 3.14)/(6.54) = 4.8 s` (ii) In this case, `omega_(0) = 0, omega = 10 xx 2 pi rad//s` ` alpha = (omega - omega_(0))/(alpha) = (20pi - 0)/(6.54) = 9.6 s` (iv) In this time, `theta = (omega^(2) - omega_(0)^(2))/(2 alpha) = ((20pi)^(2) - 0)/(2 xx 6.54) = (400pi^(2))/(13.08)` Number of revolutions `= (theta)/(2pi) = (400pi^(2))/(2pi xx 13.08) = 48.0`

Discussion

10.	A Meery -go-round, made of a ring-like plarfrom of radius `R and mass M`, is revolving with angular speed `omega`. A person of mass `M` is standing on it. At one instant, the person jumps off the round, radially awaay from the centre of the round (as see from the round). The speed of the round after wards isA. `2 omega`B. `omega`C. `omega//2`D. `0`
Answer» Correct Answer - A When the person jumps off the round, radially away from the centre, no torque is exerted i.e. `tau = 0`. According to the principle of conservation of angular momentum, `I xx omega =` constant. As mass reduces to half (from `2 M` to `M)`, moment of inertia `I` becomes half. Therefore, `omega` must become twice `(=2 omega)`.

Discussion

11.	A car is moving at a speed of `72 km//h`. The diamter of its whells is `0.5 m`. If the wheels are stopped in `20` rotations by applying brakes, calcualte the angular retardation produced by the brakes.
Answer» Here, `u = 72 km//h = (72 xx 1000)/(60 xx 60) m//s = 20m//s` `r = 0.5//2 m = 0.25 m` `omega_(2) = 0, theta = 20 xx 2 pi radian, alpha = ?` `omega_(1) = (u)/(r ) = (20)/(0.25) = 80 rad//s` From `omega_(2)^(2) - omega_(1)^(2) = 2 alpha theta` `0 - (80)^(2) = 2 alpha (20 xx 2pi)` `alpha = - (80 xx 80)/(80pi) = - 25.5 rad//s^(2)`

Discussion

12.	A thin uniform rod of length l and mass m is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is `omega`. Its cenre of mass rises to a maximum height of :A. `(I omega)/( 6 g)`B. `(l^(2)omega^(2))/(2g)`C. `(l^(2)omega^(2))/(6g)`D. `(l^(2)omega^(2))/(3 g)`
Answer» Correct Answer - C If centre of mass rises to a maximum height h, then from loss in `K.E. =` gain in `P.E.`., we get `(1)/(2)I omega^(2) = mgh` `(1)/(2)((ml^(2))/(3))omega^(2) = mgh` `h = (l^(2)omega^(2))/(6g)`

Discussion

13.	The sun rotates around itself once in`27` days. If it were to expand to twice its present diameter, what would be its new period of revolution ?
Answer» Here, `T_(1) = 27 days, R_(2) = 2R_(1), T_(2) = ?` Applying principle of conservation of angular momentum : `I_(2) omega_(2) = I_(1)omega_(1)` `((2)/(5)MR_(2)^(2)) (2pi)/(T_(2)) = ((2)/(5)MR_(1)^(2)) (2pi)/(T_(1))` `(R_(2)^(2))/(T_(2)) = (R_(1)^(2))/(T_(1))` or `T_(2) = (R_(2)^(2))/(R_(1)^(2)) xx T_(1) = (2)^(2) xx 27 = 108 days`

Discussion

14.	A solid cylinder is rolling down on an inclined plane of angle `theta`. The minimum value of the coefficient of friction between the plane and the cylinder to allow pure rolling
Answer» `mu = (1)/(3) tan theta`

Discussion

15.	A cylinder of mass `10 kg` and radius `15 cm` is rolling perfectly on a plane of inclination `30^(@)`. The coefficient of static friction is `mu_(s) = 0.25`. (a) How much is the force of friction acting on the cylinder ? (b) What is the work done against friction during rolling ? (c ) If the inclination `theta` of plane is increased, at what value of `theta` does the cylinder begin to skid and not roll perfectly ?
Answer» Here, `m = 10 kg, r = 15 cm = 0.15 m` `theta = 30^(@), mu_(s) = 0.25` Acceleration of the cylinder down the incline, `a = (2)/(3) g sin theta = (2)/(3) xx 9.8 sin 30^(@) = (9.8)/(3) m//s^(2)` (a) Force of friction, `F = mg sin theta - ma = m(g sin theta - a) = 10 (9.8 sin 30^(@) - (9.8)/(3)) = 16.4 N` (b) During rolling, the point of contact is at rest. Therefore, work done against friction is zero. (c ) For rolling without slipping/skidding, `mu = (1)/(3) tan theta` `tan theta = 3 mu = 3 xx 0.25 = 0.75` `theta = 37^(@)`

Discussion

16.	Read each statement below carefully and state with reasons, if it is true ot false. (a) During rolling the force of friction acts in the same direction as the direction of motion of c.m of the body. (b) The instantaneous speed of the point of contact during rolling is zero. (c ) The instantaneous acceleration of the point of contact during rolling is zero. (d) For perfect rolling motion, work done against friction is zero. (e) A wheel moving down a perfectly frictionless inclined plane will undergo slipping (not rolling motion).
Answer» (a) True. When a body rolls without slipping, the force of friction acts in the same direction as the direction of motion of the centre of mass of rolling body. (b) True. This is because rolling body can be imagined to be rotating about an axis passing through the point of contact of the body with the ground. Hence its instantaneous speed is zero. (c ) False. This is because when the body is rotating, its instantaneous spped is zero. (d) True. For perfect rolling motion as there is no relative motion at the point of contact, hence work done against friction is zero. (e ) True. This is because rolling occurs only on account of friction which is a tangential force capable of providing torque. When the inclined plane is perfectly smooth, it will simply slip under the effect of its own weight.

Discussion

17.	Calculate moment of inertia of a circular disc about a transverse axis through the centre of the disc. Given, diameter of disc is `40 cm`, thickness `= 7 cm` and density of material of disc `= 9 g cm^(-3)`.
Answer» Here, `R = 40//2 = 20cm, x = 7 cm, rho = 9 g cm^(-3)` Mass of disc, `M = piR^(2).x.rho` `=(22)/(7) (20)^(2) xx 7 xx 9 = 7.92 xx 10^(4) g` `I = (1)/(2)MR^(2) = (1)/(2) xx 7.92 xx 10^(4) (20)^(2)` `= 1.584 xx 10^(7) g cm^(2)`

Discussion

18.	What is the moment of inertia of a circular disc about one of its diameters ?
Answer» We assume the moment of inertia of the disc about an axis perpendicular to it and through its centre to be known, it is `MR^(2)//2`, where M is the mass of the disc and R is its radius (Table 7.1) The disc can be considered to be a planar body. Hence the theorem of perpendicular axes is applicable to it. As shown in Fig. 7.30, we take three concurrent axes through the centre of the disc, O, as the `x-, y- " and " z0` axes, `x- " and " y-` axes lie in the plane of the disc and z-axis is perpendicular to it. By the theorem of perpendicular axes. `I_(Z)=I_(X)+I_(y)` Now x and y axes are along two diameters of the disc, and by symmetry the moment of inertia of the disc is the same about any diameter. Hence `I_(x)=I_(y)` and `I_(z)=2I_(x)` But `I_(z)=MR^(2)//2` So finally, `I_(x)=I_(x)//2=MR^(2)//4` Thus the moment of inertia of a disc about any of its diameter is `MR^(2)//4`

Discussion

19.	From a circular disc of radius R and mass 9 M , a small disc of radius R/3 is removed from the disc. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O is A. `4 MR^(2)`B. `(40)/(9)MR^(2)`C. `10 MR^(2)`D. `(37)/(9)MR^(2)`
Answer» Correct Answer - A Mass per unit area of disc `= (9M)/(piR^(2))` Mass of removed portion of disc `= (9M)/(piR^(2)) xx pi ((R )/(3))^(2) = M` Moment of inertia of removed portion about an axis passing through centre of disc and perpendicular to the plane of disc, using theorem of parallel axis is `I_(1) = (M)/(2) ((R )/(3))^(2) + M ((2R)/(3))^(2) = (1)/(2)MR^(2)` when portion of disc would not have been removed, then the moment of inertia of complete disc about the given axis is `I_(2) = (1)/(2)MR^(2)` So moment of inertia of the disc with removed portion, about the given axis is `I = I_(2) - I_(1) = (9)/(2)MR^(2) - (1)/(2)MR^(2) = 4MR^(2)`

Discussion

20.	The earth has a mass of `6 xx 10^(24) kg` and a radius of `6.4 xx 10^(6)m`. Calcualte the amount of work that must be done to slow down its rotation so that duration of day becomes `30` hrs instead of 24 hours. Moment of inertia of earth `= (2)/(5)MR^(2)`.
Answer» Here, `M = 6 xx 10^(24) kg, R = 6.4 xx 10^(6)m, W =?` `T_(1) = 24 hrs, T_(2) = 30 hrs`. `I = (2)/(5)MR^(2) = (2)/(5) xx 6 xx 10^(24) xx (6.4 xx 10^(6))^(2)` `= 9.83 xx 10^(37)kg m^(2)` Work done = Decrease in rotational `KE` `W = (1)/(2)I (omega_(2)^(2) - omega_(1)^(2)) = (I)/(4) xx 4pi^(2) [(1)/(T_(2)^(2))-(1)/(T_(1)^(2))]` `=(2)/(5) MR^(2) xx 2pi^(2) ((T_(1)^(2) - T_(2)^(2))/(T_(1)^(2) T_(2)^(2)))` `= 9.83 xx 10^(37) xx 2 xx (22)/(7) xx (22)/(7)` `xx{((24 xx 60 xx 60)^(2) - (30 xx 60 xx 60)^(2))/((24 xx 60 xx 60)^(2) (30 xx 60 xx 60)^(2))}` `W =- 9.36 xx 10^(28) J`

Discussion

21.	Calculate the moment of inertia of a circular disc of radius `10 cm`, thickness `5 mm` and uniform density `8 g cm^(-3)`, about a transverse axis through the centre of the disc.
Answer» Here, `r = 10 cm, t = 5mm = 0.5 cm`, `rho = 8 g cm^(-3)` Mass of disc, `m = pi r^(2) xx r xx rho` `3.14 (10)^(2) xx 0.5 xx 8 = 1.256 xx 10^(3) gm` `I =(1)/(2)mr^(2) = (1)/(2) xx1.256 xx 10^(3) xx 10^(2)` `= 6.28 xx 10^(4) g cm^(2)`

Discussion

22.	A square of side `4 m` having uniform thickness is divided into four equal squares as shown in Fig. If one of the squares is cut off, find the position of centre of mass of the remaining portion from the centre `O`.
Answer» Let `m` be the mass of each small square. Therefore, total mass of given square `= 4m`. This acts at the centre `O` of the square. Let `O_(1)` be cm of the cut off square (shown shaded) of mass `m and O_(2)` be cm of the remaining square of mass `3m`. Now, `AC = sqrt(AB^(2) + BC^(2)) = sqrt(4^(2) + 4^(2)) = 4sqrt(2) cm` `OC = (1)/(2)AC = (4sqrt(2))/(2) = 2sqrt(2) cm` `OO_(1) =(1)/(2) = sqrt(2) cm` Now, moment of unshaded portion of mass `(3 m)` about` O =` moment of shaded portion of mass `(m)` about `O`. `:. 3m xx OO_(2) = m xx OO_(1)` `OO_(2) = (1)/(3) (OO_(1)) = (1)/(3) xx sqrt(2) cm`

Discussion

23.	Fig. shown a uniform square plate from identical squares at the corners can be removed. (a) Where is the centre of mass of the plate originally ? (b) Where is it after square 1 is removed ? (c ) where is it after squares 1 and 2 are removed ? (d) Where is `c.m` after squares `1, 2, 3,` are removed ? (f) Where is `c.m` after all the four squares are removed ? Answer in terms of quadrants and axes.
Answer» (a) Originally, centre of mass is at the centre O. (b) After square 1 is removed, `c.m` lies in quardent `3`. (c ) After squares 1 and 2 are removed c.m lies on Y-axis below O. (d) When squares 1 and 3 are removed, c.m will remain at O. (e) When squares `1, 2, 3,` are removed, c.m will shift to fouth quadrant. (f) When all the four squares are removed, c.m will shift back to O`.

Discussion

24.	Moment of inertia of a body about a given axis is the rotational inertia of the body about that axis. It is respresented by `I = MK^(2)`, where `M` is mass of body and `K` is radius of gyration of the body about that axis. It is a scalar quantity, which is measured in `kg m^(2)`. when a body rotates about a given axis, and teh axis of rotates also moves, then total `K.E.` of body `= K.E.` of translation `+ K.E.` of rotation `E = (1)/(2)mv^(2) + (1)/(2)I omega^(2)` Which the help of the compreshension given above, choose the most apporpriate altermative for each of the following questions : Kinetic energy of rotation of the flywheel in the above case isA. `20 J`B. `20 J`C. `395 J`D. `80 J`
Answer» Correct Answer - C `K.E.` of rotation `= (1)/(2)I omega^(2) = (1)/(2)I (2pin)^(2)` `= (1)/(2) xx 5 (2pi xx(120)/(60))^(2) = 395.1 J`

Discussion

25.	Moment of inertia of a body about a given axis is the rotational inertia of the body about that axis. It is respresented by `I = MK^(2)`, where `M` is mass of body and `K` is radius of gyration of the body about that axis. It is a scalar quantity, which is measured in `kg m^(2)`. when a body rotates about a given axis, and teh axis of rotates also moves, then total `K.E.` of body `= K.E.` of translation `+ K.E.` of rotation `E = (1)/(2)mv^(2) + (1)/(2)I omega^(2)` Which the help of the compreshension given above, choose the most apporpriate altermative for each of the following questions : Moment of inertia of a body depends on (i) mass of body (ii) size and shape of body (iii) axis of rotation of body (iv) all the aboveA. (i) and (ii)B. (i) and (iii)C. (ii) and (iii)D. (iv)
Answer» Correct Answer - D Moment of inertia of a body a given axis depends upon all the three factors.

Discussion

26.	Moment of inertia of a body about a given axis is the rotational inertia of the body about that axis. It is respresented by `I = MK^(2)`, where `M` is mass of body and `K` is radius of gyration of the body about that axis. It is a scalar quantity, which is measured in `kg m^(2)`. when a body rotates about a given axis, and teh axis of rotates also moves, then total `K.E.` of body `= K.E.` of translation `+ K.E.` of rotation `E = (1)/(2)mv^(2) + (1)/(2)I omega^(2)` Which the help of the compreshension given above, choose the most apporpriate altermative for each of the following questions : A circular disc and a circular ring of same mass and same diameter have, about a given axis,A. same moment of inertiaB. unequal moments of inertiaC. cannot sayD. sometimes equal sometimes not
Answer» Correct Answer - B `I_("disc") = (1)/(2)MR^(2), I_("ring") = MR^(2)` The two are unequal.

Discussion

27.	Moment of inertia of a body about a given axis is the rotational inertia of the body about that axis. It is respresented by `I = MK^(2)`, where `M` is mass of body and `K` is radius of gyration of the body about that axis. It is a scalar quantity, which is measured in `kg m^(2)`. when a body rotates about a given axis, and teh axis of rotates also moves, then total `K.E.` of body `= K.E.` of translation `+ K.E.` of rotation `E = (1)/(2)mv^(2) + (1)/(2)I omega^(2)` Which the help of the compreshension given above, choose the most apporpriate altermative for each of the following questions : A `40 kg` flywheel in the from of a unifrom circular disc of diameter `1 m` is making `120 rpm`. Its moment of inertia about a transverse axis through its centre isA. `40 kg m^(2)`B. `5 kg m^(2)`C. `10 kg m^(2)`D. `20 kg m^(2)`
Answer» Correct Answer - B `I = (1)/(2)MR^(2) = (1)/(4) xx 40 ((1)/(2))^(2) = 5 kg m^(2)`

Discussion

28.	Three identicle particle each of mass `1 kg` are placed with their centres on a straight line. Their centres are marked `A, B and C` respectively. The distance of centre of mass of the system from `A` is.A. `(PQ + PR + QR)/(3)`B. `(PQ + PR)/(3)`C. `(PQ + QR)/(3)`D. `(PQ + QR + PR)/(3)`
Answer» Correct Answer - B Clearly `CM` of the system is at `Q`, its distance from `P`, i.e. `PQ = (mxx0+mxxPQ+mxxPR)/(3m) = (PQ + PR)/(3)`

Discussion

29.	The radius of a wheel of car is `0.4m`. The car is accelerated from rest by an angular acceleration of `1.5 rad//s^(2)` for `20 s`. How much distance will the wheel cover in this time and what will be its linear velaocity ?
Answer» Here, `r = 0.4m, omega_(1) = 0`, `alpha = 1.5 rad//s^(2), t = 20 s, theta = ?`, From `theta = omega_(1) + (1)/(2) alpha t^(2) = 0 + (1)/(2) xx 1.5 (20)^(2)` `= 300 rad`. Distance covered, `s = r theta = 0.4 xx 300 m` `= 120 m` Again, `omega_(2) = omega_(1) + alpha t = 0 + 1.5 xx 20` `= 30 rad//s` `upsilon = r omega_(2) = 0.4 xx 30 = 12 m//s`

Discussion

30.	A solid cylinder of mass `20 kg` rotates about its axis with angular speed `100 s^(-1)`. The radius of the cylinder is `0.25 m`. What is the kinetic energy associated with the rotation of the cylinder ? What is the magnitude of angular momentum of the cylinder about its axis ?
Answer» M=20 kg Angular speed, `w=100 rad s^(-1), R=0.25 m` Moment of inertia of the cylinder about its axis `=1//2 MR^(2)=1//2xx20(0.25)^(2) kg m^(2)=0.625 kg m^(2)` Rotational kinetic energy, `E_(r )=1//2 lw2=1//2xx0.625xx(100)^(2)J=3125 J` Angular momentum, `L=l_(w)=0.625xx100 J_(s)=625.5 J_(s)`.

Discussion

31.	A rope of negligible mass is wound around a hollow cylinder of mass `3 kg` and radius `40 cm`. What is the angular acceleration of the cylinder, if the rope is pulled with a force of `30 N` ? What is the linear acceleration of the rope ? Assume that there is no slipping.
Answer» Here, M=3 kg, R=40 cm =0.4 m Moment of inertia of the hollow cylinder about its axis `l=MR^(2)=3(0.4)^(2)=0.48 kg m^(2)` Force applied F=30 N `therefore " Torque", tau=FxxR=300xx0.4=12 N-m`. If `alpha` is angular acceleration produced, then from `tau=I alpha` `alpha=(tau)/(I)=(12)/(0.48)=25 rad s^(-2)` Linear acceleration, `a=Ralpha=0.4xx25=10 ms^(-2)`.

Discussion

32.	A `40 kg` flywheel in the from of a unifrom circular disc `200 cm`. in diameter is making `120` revolutions/minute. Calculate angular momentum. Moment of inertia of disc `= (1//2) mass xx ("radius")^(2)`.
Answer» Here, `m = 40 kg, d = 200 cm`, `r = (d)/(2) = (200)/(2) cm = 100cm = 1m` `n = 120rps = (120)/(60)rps = 2rps, L = ?` `L = I omega = ((1)/(2)mr^(2)) 2pi n = mr^(2) pin` `= 40 xx 1^(2) xx (22)/(7) xx 2kg m^(2) s^(-1)` `L = 251.4 kg m^(2) s^(-1)`

Discussion

33.	Two discs of radii `R and 2R` are pressed against eachother. Initially, disc wit radius `R` is rotating with angular velocity `omega` and other disc is stationary. Both discs are hinged at their respective centres and are free to rotate about them. Moment of inertiaof smaller disc is `I` and of bigger disc is `2I` about their respective axis of rotation. Find the angular velocity of bigger disc after long time.
Answer» As the discs are pressed against eachother, an equal and opposite frictional force will act on both the discs, till the time rolling starts and hence the linear speed of the circumference of the two discs becomes the same. Let `omega_(1) =` angular velocity of smaller disc after long time `t` `omega_(2) =` angular velocity of bigger disc after long time `t` `f =` force of friction between the two discs when rolling starts, `upsilon_(1) = upsilon_(2)` `omega_(1)R = omega_(2)(2R) :. omega_(1) = 2omega_(2)` For smaller disc, `tau = f xx R = (I (omega - omega_(1)))/(t)` ..(i) For bigger disc, `tau_(2) = fxx 2R = (2I(omega_(2) - 0))/(t)` or `f xx R = (I omega_(2))/(t)` ..(ii) From (i) and (ii), `omega - omega_(1) = omega_(2)` or `omega - 2 omega_(2) = omega_(2)` `omega = 2 omega_(2)` or `omega = omega//3`

Discussion

34.	Two identical discs of same radius `R` are rotating about their axes in opposite directions with the same constant angular speed `omega` . The discs are in the same horizontal plane. At time `t = 0` , the points `P` and `Q` are facing each other as shown in the figure. The relative speed between the two points `P` and `Q` is `v_(r)`. In one time period `(T) ` of rotation of the discs , `v_(r)` as a function of time is best represented by A. B. C. D.
Answer» Correct Answer - A At an instant, speed of `P = v`, going in clockwise direction Speed of `Q = v`, going in anticlockwise direction Relative angular velocity of `P w.r.t Q = omega - (- omega) = 2 omega` Relative angualr separation of `P` and `Q` in time `t theta = 2 omegat`. Relative speed between the points `P` and `Q` at time `t` `\| overset rarr(upsilon_(r ))\| = sqrt(upsilon^(2) + upsilon^(2) - 2 upsilon upsilon cos (2 omegat))` `= sqrt(2 upsilon^(2) (1-cos 2 omegat)) = sqrt(2 upsilon^(2) xx 2 sin^(2) omegat)` `= 2 upsilon sin omega t` Since `\| overset rarr(upsilon_(r ))\|` will not have negative value so the lower part of the sine wave will come upper side. Hence option (a) is true.

Discussion

35.	There are two spheres of same mass and same radius, one is solid and other is hollow. Which of them has a larger moment of inertia about its dimeter ?
Answer» The hollow sphere shall have greater moment of inertia, as its entire mass is concentrated at the boundary of the sphere, at maximum distance from the axis.

Discussion

36.	Moment of inertia of a hollow cylinder of mass `M` and radius `R`, about the axis of cylinder isA. `(1)/(2)MR^(2)`B. `MR^(2)`C. `(2)/(3)MR^(2)`D. `(2)/(5)MR^(2)`
Answer» Correct Answer - B

Discussion

37.	Moment of inertia of a hollow cylinder of mass `M` and radius `R`, about the axis of cylinder is
Answer» Correct Answer - `I = mr^(2)`.

Discussion

38.	Statement-1 : A hollow cylinder of diameter `0.5 m` has a mass of `10 kg`. Its moment of inertia about its axis of symmetry is `0.625 kg m^(2)`. Statement-2 : `I = MR^(2)`A. Statement-1 is true, Statement-2 is true , Statement-2 is correct explanation of Statement-1.B. Statement-1 is true, Statement-2 is true but Statement-2 is not correct explanation of Statement-1.C. Statement-1 is true, but statement-2 is false.D. Statement-1 is false, but statement-2 is true.
Answer» Correct Answer - A From `I = MR^(2) = 10 ((0.5)/(2))^(2)` `= 0.625 kg m^(2)` Both the statements are true and statement-2 is correct explanation of statement-1.

Discussion

39.	Why does a solid sphere have smaller moment of inertia than a hollow cylinder of same mass and radius, about an axis passing through their axes of symmentry ?
Answer» By definition, `I = underset(i)sum m_(i)r_(i)^(2)` In case of hollow cylinder, entire lies at disatnce `R` from the axis of symmetry. But in case of a solid sphere, the same mass is distributed throughout at distance from zero to `R` from the axis of symmetry. Here, `R` is radius of cylinder/sphere. That is why a solid sphere has smaller moment of inertia than a hollow cylinder of same mass and radius.

Discussion

40.	A unifrom rod of length `l` and mass `m` is free to rotate in a vertical plane about `A` as shown in Fig. The rod initially in horizontal position is released. The initial angular acceleration of the rod is A. `(3g)/(2l)`B. `(3g)/(l)`C. `(g)/(2l)`D. `(2g)/(l)`
Answer» Correct Answer - A From `I alpha = tau = F xx r` `(ml^(2))/(3) alpha = mg xx (l)/(2)` `alpha = (3g)/(2l)`

Discussion

41.	A unifrom rod of length `l` and mass `m` is free to rotate in a vertical plane about `A`, Fig. The rod initially in horizontal position is released. The initial angular acceleration of the rod is `(MI "of rod about" A "is" (ml^(2))/(3))` A. `mgl//2`B. `3 g//2l`C. `2l//3g`D. `3g//2l^(2)`
Answer» Correct Answer - B Torque on rod = moment of weight of the rod about `A` `tau = mg xx (l)/(2)` As `tau = I alpha = mg (l)/(2) :. (ml^(2))/(3) alpha = mg (l)/(2)` `alpha = (3g)/(2l)`

Discussion

42.	Given the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie on the body ?
Answer» In all the four cases, as the mass density is uniform, centre of mass is located at their respective geometrical centres. No, it is not necessary that the centre of mass of a body should lie on the body. For example, in case of a circular ring, centre of mass is at the centre of the ring, where there is no mass.

Discussion

43.	Calculate moment of inertia of a uniform circular ring of mass `1.5 kg` and diameter `50 cm` about a diameter of the ring.
Answer» Here, `M = 1.5 kg, R = 50//2 = 25 cm = (1)/(4)` About a diameter of the ring. Moment of inertia, `I = (1)/(2)MR^(2) = (1)/(2)xx1.5 xx ((1)/(4))^(2) = 0.047 kg m^(2)`

Discussion

44.	Moment of inertia of a uniform circular ring of mass `2 kg` and diameter `1 m` about its diameter isA. `0.25 kg m^(2)`B. `0.5 kg m^(2)`C. `1 kg m^(2)`D. `2 kg m^(2)`
Answer» Correct Answer - A

Discussion

45.	A circular ring of radius `10 cm` is made of a wire of mass `0.02 g//cm`. Calculate its radius of gyration and moment of inertia about an axis passing perpendicular to its plane through the centre of the ring.
Answer» Here, `r = 10 cm = 0.1 m` Length of wire `= 2 pi r = 2 pi xx 10 = 20 pi cm` Mass of ring, `m = 0.02 xx 20pi kg = 0.4 pi kg` About an axis passing perpendicular to its plane, through centre of the ring, `K = r = 0.1 m` Moment of inertia of the ring about the given axis, `I = mK^(2) = mr^(2)` `= 0.4 pi (0.1)^(2) = 4pi xx 10^(-3) kg m^(2)`

Discussion

46.	If `a_(r )` and `a_(t)` respresent radial and tangential acceleration, the motion of a particle will be circualr isA. `a_(r) = 0` and `a_(t) = 0`B. `a_(r) = 0` but `a_(t) != 0`C. `a_(r) != 0` and `a_(t) = 0`D. `a_(r) !=0` and `a_(t) != 0`
Answer» Correct Answer - C::D In uniform circular motion, `a_(t) = 0`, but in non uniform circular motion, `a_(t) != 0`. In both cases, `a_(r) != 0`.

Discussion

47.	A moton car is travelling at `30 m//s` on a circular road of radius `500m`. It is increasing in speed at the rate of `2 ms^(-2)`. What is its acceleration ?
Answer» Here, `upsilon = 30m//s, r = 500m`, Tangential acceleration, `a_(t) = 2ms^(-2), a = ?` Radial acceleration, `a_(r ) = (upsilon^(2))/(r ) = (30 xx 30)/(500) = 1.8ms^(-2)` As `a = sqrt(a_(r )^(2) + a_(t)^(2))` `:. a = sqrt(1.8^(2) + 2^(2))` `= 2.7 ms^(-2)`

Discussion

48.	A sphere is rolled on a rough horizontal surface. It gradually slows down and stops. The force of friction tries toA. increase the angualr velocityB. decrease the angualr velocityC. increase the linear momentumD. decrease the linear velocity
Answer» Correct Answer - A::D Force of friction produces the torque for rotation. Therefore, it tries to increase the angualr velocity. However, friction opposes the motion. Therefore, it tries to decrease the linear velocity.

Discussion

49.	From a uniform disc of radius `R`, a circular section of radius `R//2` is cut out. The centre of the hole is at `R//2` from the centre of the original disc. Locate the centre of mass of the resulting flat body.
Answer» Let from a bigger uniform disc of radius R with centre O a smaller circular hole of radius `R//2` with its centre at `O_(1)` (where R `O O_(1)=R//2)` is cut out. Let centre of gravity or the centre of mass of remaining flat body be at `O_(2)`, where ` O O_(2)=x`. If `sigma` be mass per unit area, then mass of whole disc `M_(1)=T T R^(2) sigma` and mass of cut out part `M_(2)=pi((R )/(2))^(2) sigma=(1)/(4)piR^(2) sigma=(M_(1))/(4)` `x=(M_(1)xx(0)-M_(2)(O O_(1)))/(M_(1)-M_(2))=(0-(M_(1))/(4)xx(R )/(2))/(M_(1)-(M_(1))/(4))=-(R )/(6)` i.e., `O_(2)` is at a distnce `R//6` from centre of disc on diametrically opposite side to centre of hole.

Discussion

50.	A flywheel of mass 1 metric ton and radius `1 m` is rotating at the rate of `420` rpm. Find the constant retarding torque required to stop the wheel in `14` rotations, assuming mass to be concentrated at the rim of the wheel.
Answer» Here, `m = 1` metric ton `= 100 kg`, `r = 1m, n_(1) = (420)/(60)rps = 7 rps, tau = ?` `n_(2) = 0, theta = 14` rotations `= 14 xx 2pi` radio From `omega_(2)^(2) - omega_(1)^(2) = 2 alpha theta` `(2 pi n_(2))^(2) - (2 pi n_(1))^(2) = 2 alpha theta` `alpha = (4pi^(2)(n_(2)^(2) - n_(1)^(2)))/(2 theta)` `= (4pi^(2)(0 - 7^(2)))/(2 xx 14 xx 2pi) = (-4 xx 49 pi^(2))/(56 pi)` `alpha =- (49)/(14) xx (22)/(7) = - 11 rad//s^(2)` When mass is concemtrated at the rim, momentum of inertia of flywheel, `I = mr^(2) = 100(1)^(2) = 1000 kg m^(2)` `tau = I alpha = 1000 (-11) =- 1000 Nm` Negative sign indictes retarding torque.

Discussion

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