Question

A cylinder rolls up an inclined plane, reaches some height, and then rolls down (without slipping throughout these motions). The directions of the frictional force acting on the cylinder are-

A. up the incline while ascending and down the incline descending
B. up the incline while ascending as well as descending  
C. down the incline while ascending and up the incline while descending
D. down the incline while ascending as well as descending
Answer :   up the incline while ascending as well as descending
Solution :
Imagine the cylinder to be moving on a frictionless surface. In both the cases the acceleration of the centre of mass of the cylinder is $${\text{g}}\,{\text{sin}}\,\theta .$$   This is also the acceleration of the point of contact of the cylinder with the inclined surface. Also no torque (about the centre of cylinder) is acting on the cylinder since we assumed the surface to be frictionless and the forces acting on the cylinder is $$mg$$  and $$N$$  which pass through the centre of cylinder. Therefore the net movement of the point of contact in both the cases is in the downward direction as shown. Therefore the frictional force will act in the upward direction in both the cases.
Rotational Motion mcq solution image
Note : In general we find the acceleration of the point of contact due to translational and rotational motion and then find the net acceleration of the point of contact. The frictional force acts in the opposite direction to that of net acceleration of point of contact.

Releted MCQ Question on
Basic Physics >> Rotational Motion

Releted Question 1

A thin circular ring of mass $$M$$ and radius $$r$$ is rotating about its axis with a constant angular velocity $$\omega ,$$  Two objects, each of mass $$m,$$  are attached gently to the opposite ends of a diameter of the ring. The wheel now rotates with an angular velocity-

A. $$\frac{{\omega M}}{{\left( {M + m} \right)}}$$
B. $$\frac{{\omega \left( {M - 2m} \right)}}{{\left( {M + 2m} \right)}}$$
C. $$\frac{{\omega M}}{{\left( {M + 2m} \right)}}$$
D. $$\frac{{\omega \left( {M + 2m} \right)}}{M}$$
Releted Question 2

Two point masses of $$0.3 \,kg$$  and $$0.7 \,kg$$  are fixed at the ends of a rod of length $$1.4 \,m$$  and of negligible mass. The rod is set rotating about an axis perpendicular to its length with a uniform angular speed. The point on the rod through which the axis should pass in order that the work required for rotation of the rod is minimum, is located at a distance of-

A. $$0.42 \,m$$  from mass of $$0.3 \,kg$$
B. $$0.70 \,m$$  from mass of $$0.7 \,kg$$
C. $$0.98 \,m$$  from mass of $$0.3 \,kg$$
D. $$0.98 \,m$$  from mass of $$0.7 \,kg$$
Releted Question 3

A smooth sphere $$A$$  is moving on a frictionless horizontal plane with angular speed $$\omega $$  and centre of mass velocity $$\upsilon .$$  It collides elastically and head on with an identical sphere $$B$$  at rest. Neglect friction everywhere. After the collision, their angular speeds are $${\omega _A}$$  and $${\omega _B}$$  respectively. Then-

A. $${\omega _A} < {\omega _B}$$
B. $${\omega _A} = {\omega _B}$$
C. $${\omega _A} = \omega $$
D. $${\omega _B} = \omega $$
Releted Question 4

A disc of mass $$M$$  and radius $$R$$  is rolling with angular speed $$\omega $$  on a horizontal plane as shown in Figure. The magnitude of angular momentum of the disc about the origin $$O$$  is
Rotational Motion mcq question image

A. $$\left( {\frac{1}{2}} \right)M{R^2}\omega $$
B. $$M{R^2}\omega $$
C. $$\left( {\frac{3}{2}} \right)M{R^2}\omega $$
D. $$2M{R^2}\omega $$

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Rotational Motion


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