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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
A flywheel of moment of inertia `3 xx 10^(2) kg m^(2)` is rotating with uniform angular speed of `4.6 rad s^(-1)`. If a torque of `6.9 xx 10^(2)Nm` retards the wheel, then the time in which the wheel comes to rest isA. `1.5 s`B. `2 s`C. `0.5 s`D. `1 s` |
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Answer» Correct Answer - B Here, `I = 3 xx 10^(2) kg m^(2), omega_(1) = 4.6 rad s^(-1)` `tau = 6.9 xx 10^(2) Nm, omega_(2) = 0, t = ?` `alpha = (tau)/(I) = (-6.9 xx 10^(2))/(3 xx 10^(2)) = - 2.3 rad//s^(2)` From `omega_(2) = omega_(1) + alpha t` `0 = 4.6 - 2.3 t, t = 2 sec` |
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| 52. |
A wheel of moment of inertia `0.500 kg-m^2` and radius 20.0 cm is rotating about its axis at an angular speed of 20.0 rad/s. It picks up a stationary particle of mass 200 g at its edge. Find the new angular speed of the wheel. |
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Answer» Here, `I_(1) = 0.5 kg m^(2), r = 20 cm = (20)/(100)m, omega_(1) = 20.0 rad//s, omega_(2) = ?` As `I_(1) = (1)/(2)m_(1)r_(1)^(2) :. m_(1) = (2I_(1))/(r_(1)^(2)) = (2 xx 0.5)/((1 xx 5)^(2)) = 25 kg` Mass picked up, `m_(2) = 200g = 0.2 kg` New moment of inertia `I_(2) = (1)/(2)m_(1)r^(2) + m_(2)r^(2) = (1)/(2)(m_(1) + m_(2))r^(2) = (1)/(2) (25 + 0.2) ((1)/(5))^(2) = (25.2)/(50) kg m^(2)` According to the principle of conservation of angular momentum, `I_(2) omega_(2) = I_(1) omega_(1)` `omega_(2) = (I_(1) omega_(1))/(I_(2)) = (0.5 xx 20 xx 50)/(25.2) = 19.84 rad//s` |
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| 53. |
A wheel rotating at an angular speed of 20 rad/s ils brought to rest by a constant trouque in 4.0 secons. If the moiment of inertia of the wheel about the axis of rotation is 0.20 `kg-m^2` find the work done by the torque in the first two seconds. |
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Answer» Here, `omega_(1) = 20 rad//s, omega_(2) = 0, t = 4 sec`. `I = 0.20 kg m^(2), W = ?` `alpha = (omega_(2) - omega_(1))/(t) = (0- 20)/(4) =- 5 rad//s^(2)` Negative sign is for deceleration. Torque, `tau = I alpha = 0.20(5) = 1.0 N-m` Angle traced in first two seconds `theta = omega_(1)t + (1)/(2) alpha t^(2) = 20 (2) + (1)/(2) (-5) (2)^(2)` `= 30 radian` `:.` Work done by the torque in 1st two seconds, `W = tau theta = 1.0(30) = 30` joule |
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| 54. |
A wheel of mass `5 kg` and radius `0.40 m` is rolling on a road without sliding with angular velocity `10 rad s^-1`. The moment of ineria of the wheel about the axis of rotation is `0.65 kg m^2`. The percentage of kinetic energy of rotate in the total kinetic energy of the wheel is. |
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Answer» Here, `m = 5 kg, r = 0.4m , omega = 10 rad//s`, `I =0.65 kg m^(2) , v = r omega = 0.4 xx 10 = 4.0 m//s` Translational `KE = (1)/(2)mv^(2) = (1)/(2) xx 5 xx (4.0)^(2) = 40 J` Rotational `KE = (1)/(2)I omega^(2) = (1)/(2) xx 0.65 (10)^(2) = 32.5 J` Total `KE =` Translational `KE +` Rotational `KE` `= 40 + 32.5 = 72.5 J` `(KE "of translational")/("Total" KE) = (40)/(72.5) = 0.552 =55.2%` |
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| 55. |
Find the maximum speed at which a car can turn round a curve of `30 m` radius on a level road if coefficient of friction between the tyress and road is `0.4. Take g = 10 m//s^(2)`. |
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Answer» Here, `upsilon_(max) = ?, r = 30m, mu = 0.4` `g = 10 m//s^(2)` on a level road, centripetal force is provided by force of friction between the tyres and road. i.e., `f gt (m upsilon^(2))/(r ) or (m upsilon^(2))/(r ) lt f (= mu R = mu mg)` `:. upsilon_(max) = sqrt(mu rg) = sqrt(0.4 xx 30 xx 10)` `= sqrt(120) = 11 m//s` |
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| 56. |
A tube of length L is filled completely with an incomeressible liquid of mass M and closed at both the ends. The tube is then rotated in a horizontal plane about one of its ends with a uniform angular velocity `omega.` The force exerted by the liquid at the other end is |
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Answer» As is know, in rotational motion, angular velocity `omega` remains constant for all segments of the liquid in the tube. `:.` Force exerted by the liquid at the other end =Force required for rotational motion `F = mr omega^(2)` ….(i) Consider a small element of the liquid of length `dx` at a distance `x` from the centre, Mass of this element of the liquid `m = (M)/(L) dx , r = x` `:. dF = mr omega^(2) = ((M)/(L) dx) x omega^(2)` `F = (M)/(L) omega^(2) int_(0)^(L) x dx = (M)/(L)omega^(2) [(x^(2))/(2)]_(0)^(L)` `F = (M)/(L)omega^(2) ((L^(2))/(2)) = (1)/(2)M omega^(2)L` |
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| 57. |
A small mass attached to a string rotates on a frictionless table top as shown in Fig. If the tension in the string is increased by pulling the string causing the radius of the circular motion to decrease by a factor of 2, the kinetic energy of the mass will A. `(1)/(4)mv_(0)^(2)`B. `2mv_(0)^(2)`C. `(1)/(2)mv_(0)^(2)`D. `mv_(0)^(2)` |
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Answer» Correct Answer - B Angular momentum remains constant because the torque of tension is zero. `:. L_(i) = L_(f) or mv_(0) R = mv (R )/(2) or v = 2v_(0)` `:.` final kinetic energy `K_(f) = (1)/(2)m (2v_(0))^(2)` `= 2mv_(0)^(2)` |
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| 58. |
Assertion : At the centre of earth, a body has centre of mass , but no centre of gravity. Reason : This is because `g = 0` at the centre of earth.A. If both, Assertion and Reason are true and the Reason is the correct explanation of the Assertion.B. If both, Assertion and Reason are true but Reason is not a correct explanation of the Assertion.C. If Assertion is true but the Reason is false.D. If both, Assertion and Reason are false. |
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Answer» Correct Answer - A At the centre of earth, `g = 0`. Therefore, a body has no weight at the surface of earth , and hance no centre of gravity. But centre of mass of a body has nothing to do with gravity. Therefore, centre of mass exists. |
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| 59. |
Assertion : Torque is due to transverse componet is not perpendicular to radial component.A. If both, Assertion and Reason are true and the Reason is the correct explanation of the Assertion.B. If both, Assertion and Reason are true but Reason is not a correct explanation of the Assertion.C. If Assertion is true but the Reason is false.D. If both, Assertion and Reason are false. |
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Answer» Correct Answer - C The assertion is true, but the reason is false. |
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| 60. |
The density of a non-uniform rod of length `1m` is given by `rho (x) = a (1 + bx^(2))` where a and b are constants and `0 le x le 1`. The centre of mass of the rod will be atA. `(3 (2+ b))/(4(3+b))`B. `(4 (2+ b))/(3(3+b))`C. `(3 (3+ b))/(4(2+b))`D. `(4 (3+ b))/(3(2+b))` |
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Answer» Correct Answer - A Here, `rho (x) = a (1 + bx^(2))` When `b rarr0, rho(x) = a = consatnt` i.e., density of rod of length 1 m is constant. In that event, centre of mass of rod would lie at `0.5 m` (i.e. at centre of rod). When we try `b rarr 0` in all four given options, we find choice (a) alone gives `x = (3(2 + b))/(4(3 + b)) = (6)/(12) = 0.5`. Therefore, choice (a) is correct. |
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| 61. |
Why is the handle of a screw made wide ? |
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Answer» Turning moment of a force = force `xx` distance `(r )` from the axis of rotation. To produce a given turning moment, force required is smaller, when `r` is large. This is what happens when handle of the screw is made wide. |
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| 62. |
If a body is rotating, is it necessarily being acted upon by an external torque ? |
| Answer» No, torque is required only for producing angular acceleration. For uniform rotation, no torque is needed. | |
| 63. |
What are the units and dimensions of moment of inertia ? Is it a vector ? |
| Answer» The units of `M.I` are `kg m^(2)` and its dimensional formula is `[M^(1)L^(2)T^(0)]`. No, it is not a vector. | |
| 64. |
What is rotational analogure of mass of a body ? |
| Answer» Rotational analogue of mass of a body is moment of inertia of the body. | |
| 65. |
A constant torque is acting on a wheel. If starting from rest, the wheel makes n rotations in t seconds, show that the angular acceleration is given by `alpha = (4pi n)/(t^(2)) rad s^(-2)`. |
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Answer» Here, `omega_(0) = 0, theta = 2 pi n` radian, `t = t, alpha = ?` As `theta = omega_(0)t + (1)/(2) alpha t^(2) :. 2pi n = 0 + (1)/(2) alpha t^(2)` `alpha = (4pi n)/(t^(2)) rad s^(-2)` |
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| 66. |
The angular velocity of earth around the sun increases when it comes closer to the sun. Why ? |
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Answer» On coming closer, distance of earth from the axis through the sun decreases.`I = MK^(2)` decreases. As `I omega` is constant, `omega` increases. |
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| 67. |
(i) A person sits near the edge of a circular platform revolving with a uniform angualr speed. What will be the change in the motion of the platform ? (ii) What if the person stars moving from the edge towards the centre of the platform ? |
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Answer» (i) As `L = I omega =` constant, and `I` increases, therefore, `omega` will decrease. (ii) As the person starts moving from the edge towards the centre of platform, `I` goes on decreasing. Hence `omega` goes on increasing. |
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| 68. |
A constant torque of `200 Nm` turns a wheel of moment of inertia `50 kg m^(2)` about an axis through its centre. The angualr velocity `2` sec after staring form rest (in rad/sec) would be : |
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Answer» Correct Answer - `(8)` Here, `tau = 200 N.m, I = 50 kg m^(2)` `omega_(1) = 0, omega_(2) = ?, t = 2 s` `alpha = (tau)/(I) = (200)/(50) = 4 rad//sec^(2)` From `omega_(2) = omega_(1) + alpha t = 0 + 4 xx 2` `= 8 rad//sec`. |
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| 69. |
A wheel is subjected to uniform angualr acceleration about its axis. Initially, its angualr velocity is zero. In the first `2 sec`, it rotates through an angle `theta_(1)`, in the next `2 sec`, it rotates through an angle `theta_(2)`. The ratio of `theta_(2)//theta_(1)` isA. `1`B. `2`C. `3`D. `5` |
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Answer» Correct Answer - C From `theta = omega_(1)t + (1)/(2) alpha t^(2)`, `theta_(1) = 0 + (1)/(2) alpha (2)^(2) = 2 alpha` `theta_(1) + theta_(2) = 0 + (1)/(2) alpha(4)^(2) = 8 alpha` `theta_(2) = 8 alpha - theta_(1) = 8 alpha - 2 alpha = 6 alpha` `(theta_(2))/(theta_(1)) = (6alpha)/(2alpha) = 3` |
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| 70. |
A wheel of rdius r and moment of inertia I about its axis is fixed at top of an inclined plane of inclination `theta` as shownin figure. A string is wrapped round the wheel and its free end supports a block of mass M which can slide on the plane. INitialy, tehwheel is rotating at a speed `omega` in direction such that the block slides pu the plane. How far wil the block move before stopping? |
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Answer» Let `a` be the deceleration of the block moving up the incline. Therefore, liner deceleration of the rim of the wheel would also be a. Angular deceleration of wheel `alpha = (a)/(r )` ..(i) It `T` is tension in the string, then equation of motion of block would be `Mg sin theta - T = Ma` ...(i) Also, `tau = T xx r = I alpha` `:. T = (I alpha)/(r ) = (Ia)/(r^(2))` using (i) Put in (ii) `Mg sin theta = Ma + T = Ma + (Ia)/(r^(2))` `= a(M+ (I)/(r^(2))) = a ((Mr^(2)+ I)/(r^(2)))` `a = (Mg r^(2) sin theta)/(I + M r^(2))` Initial velocity of the block up the incline `v = r omega` `:.` Distance moved by the block before stopping `s = (v^(2))/(2a) = (r^(2)omega^(2)(I + Mr^(2)))/(2Mg r^(2) sin theta)` `s = ((I + Mr^(2)) omega^(2))/(2Mg sin theta)` |
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| 71. |
To open or close a heavy door, why force is applied at right angles to the door. |
| Answer» When `theta = 90^(@), tau = rF sin 90^(@) = rF =` maximum. | |
| 72. |
What is angular impulse ? |
| Answer» The angular impulse acting on a body is the change in angular momentum of the body about a given axis. | |
| 73. |
Find the position of centre of mass of the T-shaped plate from `O`, in Fig. |
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Answer» Suppose `sigma` is mass per unit area of the plate. `:.` Mass of horizontal portion `= 8 xx 2 sigma = 16 sigma` Mass of vertical portion `= 6 xx 2 sigma = 12 sigma` Centre of mass `O_(1)` of horizontal portion is such that `OO_(1) = 1m`. Centre of mass `O_(2)` of vertical portion is such that `OO_(2) = 2 +3 = 5m`. If `y` is distance of cm of T-shaped plate from `O`, then `y = (16sigma xx 1+12 sigma xx 5)/(16 sigma + 12 sigma) = (76 sigma)/(28 sigma) = 2.71m` |
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| 74. |
A force of `36N` is applied to a particle located at `0.15 m` from the axis of rotation. What is the magnitude of the torque about this axis, if the angle between the direction of the applied force and radius vector is (a)`120^(@) (b) 45^(@)`? |
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Answer» Here, `F = 36N, r = 0.15 m, tau = ?` (a) `theta = 120^(@)` `tau = rF sin theta` `= 0.15 xx 36 sin 120^(@) = 0.15 xx 30 xx (sqrt(3))/(2) = 4.67 N-m` (b) `theta = 45^(@)` `tau = rF sin theta` `= 0.15 xx36 sin 45^(@) = 0.15 xx 36 xx (1)/(sqrt(2)) = 3.81 N-m` |
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| 75. |
A circular ring of diameter `40 cm` and mass `1 kg` is rotating about an axis normal to its plane and passing through the centre with a frequency of `10 rps`. Calculate the angular momentum about the axis of rotation. |
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Answer» Here, `R = (40)/(2) cm = (1)/(5)m, M = 1 kg` `v = 10rps, omega = 2pi v = 2pi xx 10 = 20 pi rad//s` `L = I omega = MR^(2)omega` `= 1 xx ((1)/(5))^(2) xx 20pi = (20pi)/(25) = 2.51 kg m^(2) s^(-1)` |
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| 76. |
A torque of `10^(8)` dyne - cm is applied to a fly wheel of mass `10 kg` and radius of gyration `50 cm`. What is the resultant angular acceleration ? |
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Answer» Here, `tau = 10^(8)` dyne-cm `=(10^(8))/(10^(5)) N xx (1)/(10^(2))m = 10 N-m` `K = 50cm = (1)/(2)cm = (1)/(2)m, alpha = ? M = 10 kg` From `tau = I alpha = (MK^(2)) alpha` `alpha = (tau)/(MK^(2)) = (10)/(10 (1//2)^(2)) = 4 rad//s^(2)` |
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| 77. |
Torques of equal magnitude are applied to hollow cylinder and a solid sphere, both having the same mass and same radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. which of the two will acquire a greater angular speed after a given time ? |
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Answer» Let M be the mass and R the radius of the hollow cylinder, and also of the solid sphere. Their moments of inertia about the respective axes are `l_(1) =MR^(2) " and " l_(2)=2//5 MR^(2)` Let T be the magnitude of the torque applied to the cylinder and the sphere, producing angular accelerations `alpha_(1) " and " alpha_(2)` respectively. Then `T=l_(1) alpha_(1)=l_(2) alpha_(2)` The angular acceleration 04 produced in the sphere is larger. Hence, the sphere will acquire larger angular speed after a given time. |
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| 78. |
A thin uniform rod, pivoted at O, is rotating in the horizontal plane with constatn angular speed `omega`, as shown in the figure. At time t = 0, a small insect starts from O and moves with constant sped v, with respect to the rod towards the other end. It reaches the end of the rod at t =T and stops. The angular speed of the system remains `omega` throughout. The magnitude of the torque `(|vecpi|)` about O, as a function of time is best represented by which plot? A. B. C. D. |
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Answer» Correct Answer - B If `M` is mass of the rod, `l` is length of the rod, its moment of inertia about the end `O` of the rod `= Ml^(2)//3`.If `m` is mass of the insect and at any time `t`, the insect is at a distance `x` from `O`, its moment of inertia about `O = mx^(2)`. Therefore, moment of inertia of the system about `O` is `I = ((Ml^(2))/(3)+mx^(2))` Angular momentum of the system `L = I omega = ((Ml^(2))/(3)+mx^(2)) omega` Torque, `tau = (dL)/(dt) = (d)/(dt) ((Ml^(2))/(3)+mx^(2))omega` `= m omega (2x) (dx)/(dt) = 2m omega x. upsilon` But `x = upsilon t`. Therefore `tau = m omega (upsilon t) upsilon = 2 m omega upsilon^(2) t` Clearly, `tau prop t` At `t = T, upsilon = 0`. Therefore, `tau = 0`. Hence, the graph between `tau and t` is a stright line upto `t = T`. For `t gt T, tau = 0`. Choice (b) is correct. |
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| 79. |
if `l_(1)` is te moment of inertia of a thin rod about an axis perpendicular to its length and passing thorugh its centre of mass and `l_(2)` te moment of inertia of the ring formed by the same rod about an axis passing through the centre of mass of the ring and perpendicular tot he plane of the ring. then find the ratio `(l_(1))/(l_(2))`.A. `(pi^(2))/(3)`B. `(3)/(pi^(2))`C. `(2)/(pi^(2))`D. `(pi^(2))/(2)` |
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Answer» Correct Answer - A `I_(1) = (Ml^(2))/(12)` If `r` is radius of ring, then `l = 2 pi r, r = l//2 pi` `l_(2) = M r^(2) = M(l//2pi)^(2) = (Ml^(2))/(4pi^(2))` `(I_(1))/(I_(2)) = (Ml^(2))/(12).(4pi^(2))/(Ml^(2)) = (pi^(2))/(3)` |
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| 80. |
A solid disc and a ring, both of radius `10 cm` are placed on a horizontal table simultaneously, with initial angular speed equal to `10 pi rad//s`. Which of the two will start to roll earlier ? The coefficient of kinetic friction is `mu_(k) = 0.2`. |
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Answer» Here, initial velocity of centre of mass is zero i.e. `u = 0`. Frictional force causes the c.m to accelerate. `mu_(k) xx mg = ma :. A = mu_(k)g` …(i) As `v = u + at :. v = 0 + mu_(k) g t` ..(ii) Torque due to friction causes retardation in the initial angular speed `omega_(0)`. i.e. `mu_(k) mg xx R = - I alpha` `alpha = (-mu_(k) mg R)/(I)` ..(iii) As `omega = omega_(0) + alpha t :. omega = omega_(0) - (mu_(k) mg R t)/(I)` ...(iv) Rolling begins, when `v = R omega` From (ii) and (iv) `mu_(k) g t = R omega_(0) - (mu_(k)mg R^(2)t)/(I)` ..(v) `:. mu_(k) g t = R omega_(0) - mu_(k) g t :. omega_(0) = (2mu_(k) g t)/(R ) or t = (omega_(0)R)/(2mu_(k) g)` ..(vi) For a disc, `I = (1)/(2)mR^(2)` `:.` from (v), `mu_(k) g t = R omega_(0) - 2 mu_(k) g t` `2mu_(k) g t = R omega_(0)` `t = (omega_(0)R)/(2mu_(k) g)` ..(vii) Comparing (vi) and (vii), we find that the disc would begin to roll earlie than the ring. We can calculate the values of `t` from (vi) and (vii) using known values of `mu_(k), g, R` and `omega_(0)`. |
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| 81. |
What constant torque should be appiled to a disc of mass `10 kg` and diameter `50 cm` so that it acquires an angular velocity of `2pi` rad/s in `4 s` ? The disc is initially at rest and rotates about an axis through the centre of the disc and in a plane perpendicular to the disc. |
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Answer» Here, `tau = ? M = 10 kg`, `R = (50)/(2) cm = (1)/(4)m` `omega = 2pi rad//s, t = 4 s, omega_(0) = 0`. Moment of inertia of disc about the given axis `I = (1)/(2)MR^(2)` `alpha = (omega - omega_(0))/(t) = (2pi - 0)/(4) = (pi)/(2) rad//s^(2)` `tau = I alpha = (1)/(2) MR^(2) xx (pi)/(2)` `= (pi)/(4) xx 10 ((1)/(4))^(2) = (110)/(224) = 0.49 N-m` |
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| 82. |
How are torque and angular momentum related ? |
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Answer» `overset rarr(tau) = (d overset rarr(L))/(dt)` Torque = rate of change of angular momentum |
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| 83. |
the length of seconds hand of a clock is `10 cm`. The speed of the tip of the hand is ……… . |
| Answer» `v = r omega = r ((2pi)/(T)) = 10 xx (2pi)/(60) = (pi)/(3) cm//s`. | |
| 84. |
A flywheel is revolving with a constant angular velocity. A chip of its rim breaks and flies away. How is its angular velocity affected ? |
| Answer» As a chip breaks off, mass and hence moment of inertia of flywhell decreases. As `I omega =` constant, therefore angular velocity `omega` increases. | |
| 85. |
Explain why friction is necessary to make the disc roll (refer to `Q.28`) in the direction indicated. (a) give the direction of frictional force at `B`, and the sense of frictional torque, before perfect rolling begins. (b) What is the force of friction after perfect rolling begins ? |
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Answer» To roll a disc, we require a torque, which can be provided only by a tangential force. As force of friction is the only tangential force in this case, it is necessary. (a) As frictional force at `B` opposes the velocity of point `B`, which is to the left, the frictional force must be to the right. The sense of frictional torque will be perpendicular to the plane of the disc and outwards. (b) As frictional force at `B` decreaese the velovity of the point of contact `B` with the surface, the perfect rolling begins only when velocity of point `B` becomes zero. Also, force of friction would become zero at this stage. |
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| 86. |
Explain how is a cat able to land on its feet after a fall taking advantage of the principle of conservation of angular momentum ? |
| Answer» While falling, a cat stretches its body alongwith the tail so that its moment of inertia `(I)` increases. As no external torque is acting, `L = I omega =` constant. As I increases, `omega` decreases and it lands gently on its feet. | |
| 87. |
Speed of rotation of a body affects the radius of gyration of the body. Comment. |
| Answer» Wrong. Radius of gyration of a body is not affected by speed of the body. | |
| 88. |
When a celling fan is switched off, its angualr velocity falls to half while it makes `36` rotations. How many more rotations will it make before coming to rest ?A. `24`B. `36`C. `18`D. `12` |
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Answer» Correct Answer - D Here `theta = 36 xx 2pi, omega = (omega_(0))/(2)` As `omega^(2) = omega_(0)^(2) = 2 alpha theta`, `((omega_(0))/(2))^(2) - omega_(0)^(2) = 2 alpha (36 xx 2pi)` `- (3omega_(0)^(2))/(4) = 144pi alpha or alpha = - (omega_(0)^(2))/(192 pi)` Again. form `2 alpha theta = omega^(2) - omega_(0)^(2)`, As finally fan will stop, So `omega = 0` and `omega_(0)//2 =` Initialy angular velocity `2 (-(omega_(0)^(2))/(192pi)theta) = 0 - ((omega_(0)^(2))/(2)), theta = 24 pi` Let no. be rotations be `n` `:. 24 pi = 2pi n or n = 12` |
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| 89. |
If earth were to shrink suddenly, what would happen to the length of the day ? |
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Answer» If earth were to shrink suddenly, its radius `R` would decrease. The moment of inertia of earth `I = (2)/(5)MR^(2)` would decrease. As no external torque is acting on earth, its angular momentum `L = I omega = I(2pi)/(T)` remains constant. As I decreases, T must decrease. Hence the length of the day will decrease. |
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| 90. |
A cord is wound around the circumference of a bicycle wheel (without tyre) of diameter `1 m`. A mass of `2 kg` is tied at the end of the cord and it is allowed to fall from rest. The weight falls `2 m` in `4 s`. The axle of the wheel is horizontal and the wheel rotates which its plane vertical. if `g = 10 ms^(-2)`, what is the angular acceleration of the wheel ? |
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Answer» Here, `r = (1)/(2)m = 0.5m,m = 2 kg` `s = 2m, t = 4s, g = 10ms^(-2), alpha = ?` As `s = u t + (1)/(2) at^(2)` `:. 2 = 0 + (1)/(2) a(4)^(2) , a= (1)/(4) m//s^(2)` As `a = r alpha :. alpha = (a)/(r ) = (1//4)/(0.5) = 0.5 rad//s^(2)` |
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| 91. |
What is angular velocity of earth spinning around its own axis ? |
| Answer» `omega = (2pi)/(T) = (2pi)/(25 xx 60 xx 60) rad//s`. | |
| 92. |
The plane of the orbit of a planet can never change on its own. Why ? |
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Answer» As is known, every planet revolves arund the sun in an elliptical orbit. The centripetal force necessary for the purpose is provided by gravitational pull of the sun on the planet. As this force acts towards the centre of the sun, external torque due to this force is zero. According to the principle of conservation of angular momentum, in the absence of external torque, angular momentum `overset rarr(L)` of the planet shall remain constant in magnitude and direction. Hence, orbit of every planet will be in a fixed plane. The plane of the orbit cannot change on its own. |
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| 93. |
The concept of centre of mass proves that laws of mechanics which are true for a point mass, are equally valid for all macroscopic bodies of any shape, size, mass, etc. For example, when vector sum of external forces acting on a system of particles is zero, the velocity of centre of mass of the system will remain constant. Read the above passage and answer the following questions : (i) Is centre of mass of a body a real point ? (ii) Does centre of mass lie within the body always ? (iii) How do you justify that `overset rarr(v_(cm)) =` constant ? |
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Answer» (i) No, centre of mass of a body is not a real point, as mass of body is distributed all over the body. It is never concentrated at the centre of mass. The concept of centre of mass is only theoretical, which helps us in simplifying the calculations. (ii) No, not always, In certain cases, centre of mass of a body may be where there is no mass (iii) When `overset rarr(F_(ext)) = sum overset rarr(F_(i)) = 0, m overset rarr(a_(cm)) = 0` or `m.(d)/(dt) (overset rarr(v_(cm))) = 0 :. overset rarr(v_(cm)) =` constant |
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| 94. |
A rigid body is said to be in translational equilibrium, if it remains at rest or moving with a constant velocity in a particular direction. For this, the net external force or vector sum of all the external forces acting on the body must be zero, i.e., `oversetrarr(F) = 0 or overset rarr(F) = sum overset rarr(F_(i)) = 0` If `U` is potential energy of the body, then `F = - (dU)/(dr) = 0 or U =` constant (max. or min.) When `U =` minimum, equilibrium of body is stable, When `U` tends to increase, equilibrium is unstable, When `U` remains constant, equilibrium is neutral. Read the above passage and answer the following questions : (i) When can a bosy moving uniformly along a straight line be in equilibrium ? What is this equlilbrium called ? (ii) What is the significance of `U =` minimum in day to day life ? |
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Answer» (i) A body moving uniformly along a straight line is in equilibrium when vector sum of all the external forces acting on the body is zero. This equilibrium is called dynamic equilibrium. (ii) When `U =` minimum, i.e. potential energy of the body is minimum, the body is in stable equilibrium. In day to day life, stable equilibrium corresponds to peace of mind and potential energy `(U)` corresponds to your involvement. Thus when your involvement is minimum, you have peace of mind and stability. |
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| 95. |
A nonzero external force on a system of particles. The velocity and the acceleration of the cente of mass are found to be `v_0 and a_0` at an instant t. It is possible thatA. `v(0) = 0, a_(0) = 0`B. `v(0) != 0, a_(0) = 0`C. `v(0) = 0, a_(0) != 0`D. `v(0) != 0, a_(0) != 0` |
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Answer» Correct Answer - C::D When external force is non zero, acc. Cannot be zero. However, the centre of mass may be at rest or may be moving with some velocity. Therefore, `v_(0) = 0 or v_(0) != 0`, but `a_(0) != 0`. |
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| 96. |
To maintain a rotor at a uniform angular speed of `200 "rad s"^(-1)`, an engine needs to transmit a torque of 180 Nm. What is the power of the engine required ? |
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Answer» Here, `a=200 rad s^(-1)`, Torque, T=180 N-m Since, Power, P=Torque (T)x angular speed (w) `=180xx200=36000` watt =36 KW. |
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| 97. |
To mainntain a rotor at a uniform angular speed of `200 s^(-1)`, an engine needs to transmit a torque of `180 N-m`. What is the power of the engine required ? |
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Answer» Here, `omega = 200 s^(-1), tau 180N-m,P = ?` As `P = tau omega :. P = 180 xx 200 = 36000 "watt" = 36 kW` |
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| 98. |
Find the torque of a force `(5 hati - 2 hatj + 3 hatk)` about the origin, which acts on a particle whose position vector is `( hati - hatj + hatk)`. |
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Answer» `overset rarr(F) = (5 hati - 2 hatj + 3hatk)` and `= (hati - hatj + hatk)` `overset rarr(tau) = overset rarr(r ) xx overset rarr(F) = (hati - hatj + hatk) xx (5hati - 2hatj + 3 hatk)` `= |[hati, hatj, hatk],[1,-1,1],[5,-2,3]|` `= hati (-3 + 2) - hatj (3 - 5) + hatk(-2 + 5)` `= (-hati + 2hatj + 3hatk)` |
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| 99. |
Find the torque of a force `(5 hati - 2 hatj + 7 hatk)` about the origin, which acts on a particle whose position vector is `(2 hati - hatj + hatk)`. |
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Answer» Here, `overset rarr(tau) = ? , overset rarr(F) = (5 hati - 2hatj + 7 hatk), overset rarr(r) = (2 hati - hatj + hatk)` `overset rarr(tau) = overset rarr(r) xx overset rarr(F) = |[hati,hatj,hatk],[2,-1,1],[5,-2,7]|` `overset rarr(tau) = hati (-7 + 2) -hati(14 -5)+hatk (-4+5) or overset rarr(tau) =- 5 hati - 9hatj + hatk` |
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| 100. |
A particle of mass `2 kg` located at the position `(hati + hatj)m` has velocity `2(hati - hatj + hatk) m//s` . Its angualr momentum about Z-axis in `kg m^(2)//s` isA. `+4`B. `- 8`C. `+ 8`D. `- 4` |
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Answer» Correct Answer - B About origin, angular momentum will be `overset rarr(L) = m(overset rarr(r ) xx overset rarr(v)) = 2(hati + hatj) x 2(hati - hatj + hatk)` `= 2 |[hatj,hatj, hatk],[1,1,0],[2,-2,2]| = 4 hati - 4 hatj - 8 hatk` Component of angular mometnum about Z-axis, `L_(z) = - 8 kg m^(2)//s`. |
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