^{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.

**(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

*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.

*α*×

*10*

^{2}, so that

*α*= 20 radian per second

^{2}.

**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

*v*/2

^{2}*g*

*v*/4

^{2}*g*

*v*/2

^{2}*g*

*v*/4

^{2}*g*

*v*/

^{2}*g*

*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

*Mv*

^{2 }+ (½)

*Iω*

^{2}=

*Mgh*

*I =*(½)

*MR*

^{2}where

*R*is the radius of the cylinder. The angular velocity

*ω*=

*v/R*. Substituting these in the above equation,

*Mv*

^{2 }+ (½)×(½)

*MR*

^{2}×

*v*

^{2}/

*R*

^{2}=

*Mgh*

*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**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].

*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

*Mv*

^{2}=

*Mgh*

*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

*θ*) of the plane is greater.

**, 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]

*larger*value of

*θ*and the

*smaller*distance to be traveled, the time taken in the case of the shorter plane is smaller.

*Mv*

^{2 }+ (½)×(

^{2}/

_{5}))

*MR*

^{2}×

*v*

^{2}/

*R*

^{2}=

*Mgh*, since the moment of inertia is (

^{2}/

_{5}))

*MR*

^{2}].

**(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})

^{–1}

^{–1}

^{–1}

^{–1}

^{–1}

*Mgh*=

*M*×10×0.5 = 5

*M*joule, where

*M*is the mass of the metre stick.

*I*

*ω*

^{2}= 5

*M*

*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)]

*ML*

^{2}/3)

*ω*

^{2}= 5

*M*

*ω*=

*v/L*and

*L*= 1 metre, the above equation becomes

*v*

^{2}/6 = 5, from which

*v*= √30 = 5.4 ms

^{–1}(approximately).

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