a disc of mass m and radius r is rolling with angular speed

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. $$2M{R^2}\omega $$

Answer : $$\left( {\frac{3}{2}} \right)M{R^2}\omega $$

Solution :
KEY CONCEPT
The disc has two types of motion namely translational and rotational. Therefore there are two types of angular momentum and the total angular momentum is the vector sum of these two.
Rotational Motion mcq solution image

In this case both the angular momentum have the same direction (perpendicular to the plane of paper and away from the reader).
$$\vec L = {{\vec L}_T} + {{\vec L}_R}$$
$${L_T} = $$ angular momentum due to translational motion.
$${L_R} = $$ angular momentum due to rotational motion about $$C.M.$$
$$L = MV \times R + {I_{cm}}\omega $$
$${I_{cm}} = M.I.$$ about centre of mass C.
$$ = M\left( {R\,\omega } \right) + \frac{1}{2}M{R^2}\,\omega $$
($$v = R\omega $$ in case of rolling motion and surface at rest)
$$ = \frac{3}{2}M{R^2}\,\omega $$

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. $$\frac{{\omega \left( {M + 2m} \right)}}{M}$$

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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 $$0.7 \,kg$$

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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. $${\omega _B} = \omega $$

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Releted Question 4