In the post dated 24

^{th}January 2008, we had discussed some multiple choice questions (on circular motion and rotation) of common interest to AP Physics B as well as AP Physics C aspirants.
As promised in that post, I give below some typical questions in this section, for the benefit of those preparing for the AP Physics C examination:

**(1)**A wheel starts from rest and rotates through 1000 radians in 10 s under the action of a constant torque. If the moment of inertia of the wheel is 4 kg m

^{2}, the torque acting on the wheel is

(a) 10 Nm

(b) 20 Nm

(c) 40 Nm

(d) 80 Nm

(e) 100 Nm

The

*angular displacement*(*θ*) at the instant*t*is given by*θ = θ*

_{0}+

*ω*

_{0}

*t*+ (½)

*αt*

^{2}where

*θ*

_{0}is the initial displacement (at

*t*=0),

*ω*

_{0}is the initial angular velocity and

*α*is the angular acceleration.

Substituting the known values in the above equation, we have

1000 = 0 + 0 + (½)

*α*×*10*^{2}, so that*α*= 20 radian per second^{2}.
Torque,

**τ**=*I***= 4×20 = 80 Nm***α***(2)**A small solid cylinder rolls up along a curved surface (fig.) with an initial velocity

*v*. It will ascend up to a height ‘

*h*’ equal to

(a) 3

*v*/2^{2}*g*
(b) 3

*v*/4^{2}*g*
(c)

*v*/2^{2}*g*
(d)

*v*/4^{2}*g*
(e)3

*v*/^{2}*g*
Since the cylinder is rolling, it has translational and rotational kinetic energies. The total kinetic energy is

*K*= (½)*Mv*^{2 }+ (½)*Iω*^{2}**where***M*is the mass,*I*is the moment of inertia and*ω*is the angular velocity of the cylinder. At the highest point, the kinetic energy will be zero since the entire energy will be converted into gravitational potential energy*Mgh*where*h*is the height. So. we have
(½)

*Mv*^{2 }+ (½)*Iω*^{2}=*Mgh*
The moment of inertia of the cylinder about its own axis is

*I =*(½)*MR*^{2}where*R*is the radius of the cylinder. The angular velocity*ω*=*v/R*. Substituting these in the above equation,
(½)

*Mv*^{2 }+ (½)×(½)*MR*^{2}×*v*^{2}/*R*^{2}=*Mgh*
Or, (¾)

*Mv*^{2}=*Mgh*, from which*h*=**3***v*/4^{2}*g***Now, suppose we change the above question as follows:**

**A small cylinder**

*rolling*with a velocity*v*along a horizontal surface encounters a*smooth*inclined surface. The height ‘*h*’*up to which the cylinder***will ascend is**

**(a) 3**

*v*/2^{2}*g***(b)**

**3**

*v*/4^{2}*g***(c)**

*v*/2^{2}*g***(d)**

*v*/4^{2}*g***(e)3**

*v*/^{2}*g*
You should remember that a body can

*roll*along a surface only if the surface is rough (so that there is frictional force). On a smooth surface the body can slide; but it cannot have a linear displacement by rolling. [You might have seen how a car tyre rotates in mud without producing any movement of the car].
In the present problem, the body will

*roll*up to the foot of the inclined smooth surface. It will continue to spin with the angular speed it has acquired, and will slide up to a certain height, maintaining its spin motion throughout the smooth surface. Its translational kinetic energy alone is responsible for its upward motion along the smooth incline so that the height up to which it will rise is given by
(½)

*Mv*^{2}=*Mgh*
Therefore,

*h = v*^{2}/2*g.***(3)**Two inclined planes have same height but different lengths (and therefore different angles). If a solid sphere is allowed to roll down from the top of these inclined planes

(a) the time of descent will be same, but the speed at the bottom of the plane will be greater for the longer plane

(b) the time of descent will be same, but the speed at the bottom of the plane will be smaller for the longer plane

(c) the time of descent and the speed at the bottom will be same in both cases

(d) time of descent will be different, but the speed at the bottom will be same in both cases

(e) speeds and the times of descent will be different

A body rolling down an inclined plane has greater acceleration if the angle (

*θ*) of the plane is greater.
[The expression for the acceleration is

**, as given in he post dated 20***a = gsinθ**/*[1*+*(*k*^{2}*/R*)]^{2}^{th}January 2008, but you can solve this problem even without remembering this equation]
The length of the steeper inclined plane is smaller since the planes have the same height. Because of the

*larger*value of*θ*and the*smaller*distance to be traveled, the time taken in the case of the shorter plane is smaller.
The velocity of the sphere at the bottom of the plane is determined by the height of the plane since the kinetic energy at the bottom is equal to the gravitational potential energy at the top, as we have seen in the previous question.

[In the case of a sphere, the equation will be (½)

*Mv*^{2 }+ (½)×(^{2}/_{5}))*MR*^{2}×*v*^{2}/*R*^{2}=*Mgh*, since the moment of inertia is (^{2}/_{5}))*MR*^{2}].
Since the height is the same, the speed at the bottom will be the same. Therefore, the correct option is (d).

**(4)**A metre stick AB hinged (without friction) at the end A as shown in the figure is kept horizontal by means of a string tied to the end B, the other end of the string being tied to a hook. The string is carefully cut (or burnt), and the scale executes angular oscillations about an axis passing through the end A. What is the speed of the end B when the metre stick assumes vertical position immediately after the string is burnt? (Acceleration due to gravity = 10 ms

^{–2})

(a) 2.5 ms

^{–1}
(b) 4.2 ms

^{–1}
(c) 5.4 ms

^{–1}
(d) 6.8 ms

^{–1}
(e) 9.8ms

^{–1}
The centre of gravity of the metre stick is at its middle. When the string is burnt, the height of the centre of gravity is reduced by 0.5 m, thereby reducing the gravitational potential energy of the metre stick by

*Mgh*=*M*×10×0.5 = 5*M*joule, where*M*is the mass of the metre stick.
The decrease in the potential energy is equal to the increase in kinetic energy. Therefore in the vertical position of the metre stick, its kinetic energy

(½)

*I**ω*^{2}= 5*M*
Since the moment of inertia of a uniform bar of length

*L*about a normal axis through ite mid point is*ML*^{2}/12, its moment of inertia about a parallel axis through its end is*ML*^{2}/3. [You will get this by applying the parallel axis theorem (see the post dated 2o^{th}January 2008)]
Therefore, we have (½)(

*ML*^{2}/3)*ω*^{2}= 5*M*
Since

*ω*=*v/L*and*L*= 1 metre, the above equation becomes*v*

^{2}/6 = 5, from which

*v*= √30 = 5.4 ms

^{–1}(approximately).

You will find some useful posts in this section at

__http://www.physicsplus.in__
this is helpful :)

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