Question

Moment of inertia of a uniform circular disc about a diameter is $$I.$$ Its moment of inertia about an axis perpendicular to its plane and passing through a point on its rim will be

A. $$5I$$
B. $$3I$$
C. $$6I$$  
D. $$4I$$
Answer :   $$6I$$
Solution :
Problem Solving Strategy
For this type of problem, always apply parallel and perpendicular axes theorem of moment of inertia. Moment of inertia of uniform circular disc about its diameter $$= I$$
According to theorem of perpendicular axes, Moment of inertia of disc about its axis $$= 2I$$
Applying theorem of parallel axes,
Moment of inertia of disc about the given axis $$ = 2I + m{r^2}$$
$$\eqalign{ & = 2I + 4I\,\,\left( {{\text{as}}\,\,2I = \frac{1}{2}m{r^2}\,\therefore m{r^2} = 4I} \right) \cr & = 6I \cr} $$

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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