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

1.	Calculate the velocity with which a solid cylinder rolls down an inclined plane of height `10m` with slipping (take `g=10m//s^(2)`)
Answer» `M.I.` of solid cylinder `=(1)/(2)MR^(2)` `K=(1)/(sqrt(2))R` `v=sqrt((2xx10xx10)/((1+(1)/(2))))=sqrt((2xx10xx10xx2)/(3))=(20)/(sqrt(3))m//s` Hint : `v=sqrt((2gh)/((1+(k^(2))/(R^(2)))))`

Discussion

2.	A sphere and a cylinder have the same radius and the same mass. They start from rest at the top of an incline. Which reaches the bottom first? Which has the greater speed at the bottom? Which has the greater total kinetic energy of the bottom ? Which has the greater rotational `KE`?
Answer» Sphere will reach first, sphere will have greater speed, both will have same kinetic energy, cylinder have greater rotational kinetic energy.

Discussion

3.	A uniform ring rolls down an inclined plane without slipping. If it reaches the bottom of a speed of `2m//s`, then calculate the height of the inclined plane (use `g=10m//s^(2)`)
Answer» `h=(v^(2))/(2g)(1+(k^(2))/(R^(2)))` `M.I.` of ring `=MR^(2)`, `K=R` `h=(v^(2))/(2g)(1+(k^(2))/(R^(2)))=(v^(2))/(2g)xx2=(v^(2))/(g)=(2xx2)/(10)=0.4m` Hint : `h=(v^(2))/(2g)(1+(k^(2))/(R^(2)))`

Discussion

4.	A torque of `5Nm` is applied on a particle for `2` seconds, calculate the change in its angular momentum.
Answer» `(dL)/(dt)=vectau` `impliesdL=2xx5=10kgm^(2)//s` Hint. `vectau=(dvecL)/(dt)`

Discussion

5.	What is the angular acceleration of a particle moving with constant angular velocity ?
Answer» Zero, `omega=` constant `(d omega)/(dt)=0`

Discussion

6.	What is the angular acceleration of a particle if the angular velocity of a particle becomes `4` times of its initial angular velocity `1` rad s in `2` seconds
Answer» Using `alpha=(omega-omega_(0))/(t)` `alpha=(4-1)/(2)=(3)/(2)rad//s^(2)` Hint : `alpha = (omega - omega_(0))/(t)`

Discussion

7.	What is the angular momentum of a particle at rest ? Hint. `vecv=0`.
Answer» Zero Hint. `vecv=0`.

Discussion

8.	A wheel is rotating with an angular velocity of `3 rad s^(-1)`. If the angular acceleration is `2 rad s^(-1)` then calculate its angular velocity after `5` second.
Answer» `alpha=(omega-omega_(0))/(t)` `implies2=(omega-3)/(5)` `impliesomega=10+3=13rads^(-1)` Hint. `alpha=(omega-omega_(0))/(t)`

Discussion

9.	A particle of mass `1kg` is moving about a circle of radius `1m` with a speed of `1m//s`. Calculate the angular momentum of the particle.
Answer» `L=mvr=1xx1xx1=1kg m^(2)s^(-1)`

Discussion

10.	A force `(hati-2hatj+3hatk)` acts on a particle of position vector `(3hati+2hatj+hatk)`. Calculate the `hatj^(th)` component of the torque acting on the particle.
Answer» `vectau=vecrxxvecF` `=\|{:(hati,hatj,hatk),(3,2,1),(1,-2,3):}\|=hati(6+2)+hatj(1-9)hatk(-6-2)` `=8hati-8hatj-8hatk` `=8(hati-hatj-hatk)` The `hatj^(th)` component of the torque is `-8`

Discussion

11.	A force `(hati-2hatj+3hatk)` acts on a particle lying at origin. What is the torque acting on the particle.
Answer» Zero, `vectau=vec rxx vecF`, `vec r=vec0` `:.vectau=vec0` Hint. `vecr=vec0`

Discussion

12.	Find a vector perpendicular to vector `vecA=(hati+2hatj-3hatk)` as well as `vecB=(hati+hatj-hatk)`
Answer» `vecC=vecAxxvecB=\|{:(hati,hatj,hatk),(1,2,-3),(1,1,-1):}\|=hati(-2+3)+hatj(-3+1)+hatk(1-2)` `=hati-2hatj-hatk` Hint. `vecC=vecAxxvecB`

Discussion

13.	Find a unit vector perpendicular to each of the vectors `hati+2hatj-3hatk` and `hati-2hatj+hatk`.
Answer» Let `vecA=hati+2hatj-3hatk` `vecB=hati-2hatj+hatk` and let `vecC` is the vector given by the cross producty of `vecA`, `vecB` and `vecC` is perpendicular to the plane containing `vecA` and `vecB` `vecC=vecAxxvecB` Using Determinant method `\|{:(hati,hatj,hatk),(1,-2,-3),(1,-2,1):}\|=hati(-9+8)-hatj(6-4)+hatk(-4+3)=(-hati-2hatj-hatk)`

Discussion

14.	A force `(hati-2hatj+3hatk)` acts on a particle of positive vector `(3hati+2hatj+hatk)`. Calculat the torque acting on the particle.
Answer» `vectau=vecrxxvecF`. using Determinant method `vectau=\|{:(hati,hatj,hatk),(3,2,1),(1,-2,3):}\|` `impliesvectau=8(hati-hatj-hatk)`Nm.

Discussion

15.	A car of mass 300 kg is travelling on a circular track of radius 100 m with a constant speed of 60 m/s Calculate the angular momentum ?
Answer» `vecL=vecrxxvecP` `vecP=mv=300xx60=18000kg m//s` (directed along the tangent to the circle) `vecL=rP sin theta`, `theta=90^(@)` `vecL=rP` `=100xx18000` `=18xx10^(5)kg ms^(-1)`

Discussion

16.	Will the velocity and acceleration of centre of mass change if particles `A`, `B` and `C` are projected from different locations but with same velocities ?
Answer» Correct Answer - No

Discussion

17.	A uniform stick of mass `M` and length `L` is kept vertical on a frictionless horizontal floor. If the stick is allowed to fall, then mark the correct alternativesA. Path of its centre of mass is ellipticalB. Path of the centre of mass of the stick is vertical straight lineC. Poin tof contact of the stick with the ground will move in parabolic pathD. Point of contact of the stick with the ground will move in straight line
Answer» Correct Answer - B,D

Discussion

18.	Chosse the correct altarnativesA. For a general rotational motion , angular momentum L and angular velocity `omega` need not be parallel .B. For a rotional motion about a fixed axis , angular momentum L and angular velocity `oemga` are always parallel .C. For a gereral translational motion , momentum P and velocity V are always parallel .D. For a general transational motion , acceleration a and velocity V are always parallel.
Answer» Correct Answer - a,c For a general rotational motion , where axis of rotation is not Angular momentum L and angular velocity `omega ` need not be parallel , for a general translational motion momentum p =mv hence , p and v are always parallel .

Discussion

19.	Figure shows two identical particles `1` and `2`, each of mass `m`, moving in opposite directions with same speed `vec V` along parallel lines. At a particular instant, `vec r_1` and `vec r_2` are their respective position vectors drawn from point `A` which is in the plane of the parallel lines. Which of the following is the correct statement ? .A. Agular momentum `I_(1)` of particle a about A is `I = mv (d_(1))o.`B. Angular momentum `I_(2)` of particle 2 about A is ` I_(2)=mvr_(2) o.`C. total angular momentum of the system about A is ` I = mv(r_(1)+r_(2))o.`D. total angular momentum of the system about A is `I=mv(d_(2)-d_(1))ox`
Answer» Correct Answer - a,b the angular momentum L of a particle with respect toorigin is deined to be `L=rxxp` where , r is the podsition vector of the particle and p is the linear momentum , The direction of L is is perpendicular to both d r and p by right hand rule . For particle 1. `I_(1) =r_(1)xxmv` is not of plane of the paper and perpendicular to `r_(1)` and p(mv) Similarly `I_(2)=r_(2)xxm(-v)` is into the plane of the paper and perpendicular to `r_(2) and -p` Hence , total angular momentum `l=l_(1)+l_(2)=r_(1)xxmv+(-r_(2)xxmv)` `\|l\|=mvd_(1)- mvd_(2)as d_(2)gt d_(1)` total angular momentum will be inward Hence `I=mv(d_(2)-d_(1))ox` note : in the expression of angular momentum `I=rxxp` the direction of l is taken by right hand rule .

Discussion

20.	Calculate the radius of gyration of a circular disc about its diameter.
Answer» `I=MK^(2)`, for disc `I=(MR^(2))/(4)` Comparing , `K=(R )/(2)` Hint : `I_("disc") = (MR^(2))/(4)` (abour an axis passing through its diameter)

Discussion

21.	The moment of inertia of a solid cylindrer about its axis of ratation is `(MR^(2))/(2)`. What is the value if the radius of gyration of the cylinder about this axis ?
Answer» `I=(MR^(2))/(2)`……..`(i)` `I=MK^(2)`…….`(ii)` Comparing `(i)`, `(ii)` `K^(2)=(R^(2))/(2)`, `K=(R )/(sqrt(2))m`

Discussion

22.	The moment of inertia of a uniform ring about an axis passing through its centre and perpendicular to its plane is `100kgm^(2)`. What is the moment of inertia of the ring about its diameter ?
Answer» Moment of inertia of the ring through `CM=100kgm^(2)` `MR^(2)=100` `I` about diameter `=(MR^(2))/(2)=50kgm^(2)`

Discussion

23.	Calculate the moment of inertia of a ring of mass `2kg` and radius `2cm` about an axis passing through its centre and perpendicular to its surface.
Answer» `I=MR^(2)` `=(2xx2)/(100)xx(2)/(100)=8xx10^(-4)kg m^(2)`

Discussion

24.	The moment of inerta of a ring of mass `1kg` about an axis passing through its centre perpendicular to its surface is `4kgm^(2)`. Calculate the radius of the ring.
Answer» `I=MR^(2)`, `R=sqrt((I)/(M))=sqrt((4)/(1))=2m`

Discussion

25.	The moment of inertia of a body moving with angular velocity `omega` decreases from `l` to `(l)/(3)` without any external torque. Calculate the new angular velocity of the body.
Answer» `I_(1)omega_(1)=I_(2)omega_(2)` `omega_(2)=(I_(1)omega_(1))/(I_(2))=(Iomega)/((I)/(3)))=3omgea` Hint : `I_(1)omega_(1)=I_(2)omega_(2)`

Discussion

26.	Calculate the angular velocity with which a wheel is rotating. If its instanteous power is `500W`, and an external torque of `50Nm` is applied to it. Hint : `P=tau omega`
Answer» `P=tau omega=50xxalphha=500` `alpha=10rad//s^(2)` Hint : `P = tau omega`

Discussion

27.	Calculate the instantaneous power of a body rotating with angular velocity of `20rad//s^(-2)`, when an external torque of `5 Nm` is applied to it. Hint : `P=tau omega`.
Answer» `P=tau omega`, `P=5xx20=100W` `P tau omega`

Discussion

28.	the density of a non-uniform rod of length 1 m is given by `rho(x)=a(1+bx^(2))` where ,a and b are constants and `0lexle1,` the centre of mass of the rod will be at.A. `(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))`
Answer» Correct Answer - a density is given as `rho (x)=a(1+bx^(2))` where a and b are constants and `0 le xle 1.` Let `b to 0,` in this case ` rho (x)=a=` constant Hence ,centre of mass will be at `x=0.5 m` (middle of the rod) Putting , b=0 in the options , only (A) given 0.5 Note :- We should not check options by puting a = 0 , because ` rho =0` for `a =0`

Discussion

29.	the variaton of angular postion `theta` of a point on a rotating rigid body with time t is shown in figure Is the body rotating clokwise or anti- clokwise ?
Answer» AS the slope of `theta` -t graph is possitive and slope indicates , anti - clock wise rotation which is traditionally taken as positive .

Discussion

30.	Give an example of a body in rotational equilibrium but not in translational equilibrium.
Answer» If the forces act in the same direction.

Discussion

31.	If the parallel forces acting on a lever are in the ratio `3 : 5` then what is the mechanical advantage of the lever ?
Answer» Machanical Advantage `=(F_(1))/(F_(2))=(3)/(5)` Hint : `M.A = (F_(1))/(F_(2))`

Discussion

32.	What is the torque acting on a body moving with constant angular velocity ?
Answer» As `omega=` constant `alpha=0` Hint : `alpha=(d omega)/(dt)`, `omega` is constant

Discussion

33.	Give conditions for mechanical equilibrium.
Answer» `(i)` Total force acting on the body should be zero. `(ii)` Total torque acting on the body should be zero.

Discussion

34.	Define the centre of gravity of a homogenous rigid body.
Answer» It is the point on the body where the whole weight of the body is supposed to be concentrated.

Discussion

35.	Find the acceleration of the pulley after `5s`, if the system is released from rest.
Answer» Correct Answer - `12.5rad//s`

Discussion

36.	Find these forces ?
Answer» Correct Answer - `45N`

Discussion

37.	A uniform square plate has a samll piace Q of an irregular shape removed and lued to the centre of the plate leving a hole behind in figure the moment if inertia about the z- axis is then , A. increasedB. decreasedC. the sameD. changed in unpedicted manner
Answer» Correct Answer - b in the given diagrams , when the small piece Q removed and glued to the centre of the plate , the mass comes closer to the z- axis , hence , moment of inertia decreases.

Discussion

38.	A wheel in uniform motion about an axis passing through its centre and perpendicular plus is considered to be in mechanical (translational plus rotational ) equilibrium because no net external force or torqure is reqired to sustain its motion However , the particles than constitute the wheel do experience a centripeteal the acceleration wheel being in equilibrium ? how would you set a half wheel into unifrom motion about an axis passing throgh the centre of mass of the wheel and perpendicular to its plane ? will ypu require external forces to sustain the motion ?
Answer» Wheel is a rigid body , the particles that consiture the wheel do experience a centripetal acceleration directed towards the centre , this acceleration arises due to internal eleastic forces , which cancel out in pairs. In a half wheel the distribution of mass about its centre of mass (through which axis of rotation passes ) is not symmetrical , therefore , the direction of angular momentum of the wheel does not coincide with the direction of its angular velocity Hence an external torque is required to maintain the motion of the half wheel.

Discussion

39.	Two particles of a mass 2m and m are tied with an inextensible string the particle of mass m is given a speed V as shown in the figure. Find the speed with which the particle of mass 2m starts moving.
Answer» Correct Answer - `(2v_(0))/(19)`

Discussion

40.	Figure shows a right angled triangle of uniform mass per unit area, attached with rectangular lamina of sides `3m` and `5m` made of same material. Find the centre of mass of the system.
Answer» Correct Answer - `2.433m`, `2.157m`

Discussion

41.	A small metallic spherical ball is dropped from a height `h`. If it collides with the ground inelastically and rebounds, thenA. The momentum of the ball just after the collision is same as that just before the collisionB. Force on the ball is same (in magnitude) as the force on the ground during the collisionC. Total momentum of ball and Earth system is conservedD. Total kinetic energy of the ball and the earth remains the same
Answer» Correct Answer - BC

Discussion

42.	A particle of mass `m` moving with a speed `u` strikes a smooth horizontal surface at an angle `45^(@)`. The particle rebounds at an angle `phi` with speed `v`. If coefficient of restituion is `(1)/(sqrt(3))`, then angle `phi` is A. `30^(@)`B. `45^(@)`C. `60^(@)`D. `37^(@)`
Answer» Correct Answer - option 1

Discussion

43.	A particle of mass `m` moving with speed `u` hits a wedge of mass `M` as shown in figure. If coefficient of restitution is `e` and friciton can be neglected, then find the speed of the wedge with respect to ground just after the collision.
Answer» `((1+e)m*usin^(2)theta)/(M+msin^(2)theta)`

Discussion

44.	A particle of mass `m` is moving anticlockwise, in a circle of radius `R` in `x-y` plane with centre at `(R,0)` with a constant speed `v_(2)`. If is located at point `(2R,0)` at time `t=0`. A man starts moving with a velocity `v_(1)` along the positive `y`-axis from origin at `t=0`. Calculate the linear momentum of the particle w.r.t. man as a function of time.
Answer» `m[(-v_(2)sin.(v_(2))/(R )t)veci+(v_(2)cos.(v_(2))/(R )t-v_(1))vecj]`

Discussion

45.	A particle is moving along a circle of radius `20cm`, with a linear velocity of `2m//s. Calculate the angular velocity of the particle.
Answer» `vecv=vecomegaxxvecr` as the angle between `vecr` and `vecomega` is `90^(@)`, so the magnitude of linear velocity is `\|vecv\|=r omega` `omega=(v)/(r )=(2xx100)/(20)` `=10"rad"s^(-1)`

Discussion

46.	The angular velocity of a rigid body is `24` rad `s^(-1)`, Calculate the time it will take to rotate `72` rad.
Answer» `omega=(theta)/(t)impliest=(theta)/(omega)=(72)/(24)=3s`

Discussion

47.	In the figure given below, what is the equilibrium of the rod i.e., is it translational or rotational ?
Answer» As the forces are acting in opposite directions therefore `sumF=0` hence the rod is in translational equilibrium.

Discussion

48.	If the Mechanical Advantage of a lever of `100`, when a force of `10N` acts as a distance of `20cm` from the point of fulcrum. Calculate the force acting on the other side of the fulcrum and the distance of that force from the fulcrum.
Answer» `MA=(d_(2))/(d_(1))=(F_(1))/(F_(2))` `implies100=(10)/(F_(2))` `impliesF_(2)=(10)/(100)=(1)/(10)N` and `(d_(2))/(d_(1))=MA` `impliesd_(2)=Maxxd_(1)` `=100xx10` `=1000cm`

Discussion

49.	The angular speed of a motor wheel is increased from `600"rpm"` to `1200"rpm"` in `4` seconds. Calculate how many revolutions does the engine make during this time.
Answer» Using `omega=omega_(0)+alpha t` `alpha=(omega-omega_(0))/(t)` `omega_(0)=600"rpm"` `=(2pixx600)/(60)=20pi rad s^(-1)` `omega=1200"rpm"` `=(2pixx1200)/(60)=40pi rad s^(-1)` `alpha=(20pi)/(4)=5pi rad s^(-2)` again using, `theta=omega_(0)t+(1)/(2)alpha t^(2)` `theta=20pixx4+(1)/(2)xx5pixx16` `theta=80pi+40pi=120pi` Number of revolutions `=(theta)/(2pi)=(120pi)/(2pi)=60` revolutions

Discussion

50.	Which of the following points is the likely position of the centre of mass of the system shown in Fig. A. AB. BC. CD. D
Answer» Correct Answer - c Centre of mass of a system lies towards the part of the system , having bigger mass , in the above diagram . Lower part is hwavier , hence CM of the system lies below the horizontal diameter.

Discussion

Explore topic-wise InterviewSolutions in .

Calculate the velocity with which a solid cylinder rolls down an inclined plane of height `10m` with slipping (take `g=10m//s^(2)`)

A sphere and a cylinder have the same radius and the same mass. They start from rest at the top of an incline. Which reaches the bottom first? Which has the greater speed at the bottom? Which has the greater total kinetic energy of the bottom ? Which has the greater rotational `KE`?

A uniform ring rolls down an inclined plane without slipping. If it reaches the bottom of a speed of `2m//s`, then calculate the height of the inclined plane (use `g=10m//s^(2)`)

A torque of `5Nm` is applied on a particle for `2` seconds, calculate the change in its angular momentum.

What is the angular acceleration of a particle moving with constant angular velocity ?

What is the angular acceleration of a particle if the angular velocity of a particle becomes `4` times of its initial angular velocity `1` rad s in `2` seconds

What is the angular momentum of a particle at rest ? Hint. `vecv=0`.

A wheel is rotating with an angular velocity of `3 rad s^(-1)`. If the angular acceleration is `2 rad s^(-1)` then calculate its angular velocity after `5` second.

A particle of mass `1kg` is moving about a circle of radius `1m` with a speed of `1m//s`. Calculate the angular momentum of the particle.

A force `(hati-2hatj+3hatk)` acts on a particle of position vector `(3hati+2hatj+hatk)`. Calculate the `hatj^(th)` component of the torque acting on the particle.

A force `(hati-2hatj+3hatk)` acts on a particle lying at origin. What is the torque acting on the particle.

Find a vector perpendicular to vector `vecA=(hati+2hatj-3hatk)` as well as `vecB=(hati+hatj-hatk)`

Find a unit vector perpendicular to each of the vectors `hati+2hatj-3hatk` and `hati-2hatj+hatk`.

A force `(hati-2hatj+3hatk)` acts on a particle of positive vector `(3hati+2hatj+hatk)`. Calculat the torque acting on the particle.

A car of mass 300 kg is travelling on a circular track of radius 100 m with a constant speed of 60 m/s Calculate the angular momentum ?

Will the velocity and acceleration of centre of mass change if particles `A`, `B` and `C` are projected from different locations but with same velocities ?

Calculate the radius of gyration of a circular disc about its diameter.

The moment of inertia of a solid cylindrer about its axis of ratation is `(MR^(2))/(2)`. What is the value if the radius of gyration of the cylinder about this axis ?

The moment of inertia of a uniform ring about an axis passing through its centre and perpendicular to its plane is `100kgm^(2)`. What is the moment of inertia of the ring about its diameter ?

Calculate the moment of inertia of a ring of mass `2kg` and radius `2cm` about an axis passing through its centre and perpendicular to its surface.

The moment of inerta of a ring of mass `1kg` about an axis passing through its centre perpendicular to its surface is `4kgm^(2)`. Calculate the radius of the ring.

The moment of inertia of a body moving with angular velocity `omega` decreases from `l` to `(l)/(3)` without any external torque. Calculate the new angular velocity of the body.

Calculate the angular velocity with which a wheel is rotating. If its instanteous power is `500W`, and an external torque of `50Nm` is applied to it. Hint : `P=tau omega`

Calculate the instantaneous power of a body rotating with angular velocity of `20rad//s^(-2)`, when an external torque of `5 Nm` is applied to it. Hint : `P=tau omega`.

the density of a non-uniform rod of length 1 m is given by `rho(x)=a(1+bx^(2))` where ,a and b are constants and `0lexle1,` the centre of mass of the rod will be at.A. `(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))`

the variaton of angular postion `theta` of a point on a rotating rigid body with time t is shown in figure Is the body rotating clokwise or anti- clokwise ?

Give an example of a body in rotational equilibrium but not in translational equilibrium.

If the parallel forces acting on a lever are in the ratio `3 : 5` then what is the mechanical advantage of the lever ?

What is the torque acting on a body moving with constant angular velocity ?

Give conditions for mechanical equilibrium.

Define the centre of gravity of a homogenous rigid body.

Find the acceleration of the pulley after `5s`, if the system is released from rest.

Find these forces ?

A uniform square plate has a samll piace Q of an irregular shape removed and lued to the centre of the plate leving a hole behind in figure the moment if inertia about the z- axis is then , A. increasedB. decreasedC. the sameD. changed in unpedicted manner

Two particles of a mass 2m and m are tied with an inextensible string the particle of mass m is given a speed V as shown in the figure. Find the speed with which the particle of mass 2m starts moving.

Figure shows a right angled triangle of uniform mass per unit area, attached with rectangular lamina of sides `3m` and `5m` made of same material. Find the centre of mass of the system.

A particle of mass `m` moving with a speed `u` strikes a smooth horizontal surface at an angle `45^(@)`. The particle rebounds at an angle `phi` with speed `v`. If coefficient of restituion is `(1)/(sqrt(3))`, then angle `phi` is A. `30^(@)`B. `45^(@)`C. `60^(@)`D. `37^(@)`

A particle of mass `m` moving with speed `u` hits a wedge of mass `M` as shown in figure. If coefficient of restitution is `e` and friciton can be neglected, then find the speed of the wedge with respect to ground just after the collision.

A particle is moving along a circle of radius `20cm`, with a linear velocity of `2m//s. Calculate the angular velocity of the particle.

The angular velocity of a rigid body is `24` rad `s^(-1)`, Calculate the time it will take to rotate `72` rad.

In the figure given below, what is the equilibrium of the rod i.e., is it translational or rotational ?

If the Mechanical Advantage of a lever of `100`, when a force of `10N` acts as a distance of `20cm` from the point of fulcrum. Calculate the force acting on the other side of the fulcrum and the distance of that force from the fulcrum.

The angular speed of a motor wheel is increased from `600"rpm"` to `1200"rpm"` in `4` seconds. Calculate how many revolutions does the engine make during this time.

Which of the following points is the likely position of the centre of mass of the system shown in Fig. A. AB. BC. CD. D