Question

Look at the drawing given in the figure which has been drawn with ink of uniform line-thickness. The mass of ink used to draw each of the two inner circles, and each of the two line segments is $$m.$$  The mass of the ink used to draw the outer circle is $$6 \,m.$$
The coordinates of the centres of the different parts are: outer circle $$\left( {0,0} \right),$$  left inner circle $$\left( { - a,a} \right),$$  right inner circle $$\left( {a,a} \right),$$  vertical line $$\left( {0,0} \right)$$  and horizontal line $$\left( {0, - a} \right).$$  The $$y$$-coordinate of the centre of mass of the ink in this drawing is
Rotational Motion mcq question image

A. $$\frac{a}{{10}}$$  
B. $$\frac{a}{{8}}$$
C. $$\frac{a}{{12}}$$
D. $$\frac{a}{{3}}$$
Answer :   $$\frac{a}{{10}}$$
Solution :
The system is made up of five bodies (three circles and two straight lines) of uniform mass distribution. Therefore we assume the system to be made up of five point masses where the mass of each body is considered at its geometrical centre.
Rotational Motion mcq solution image
The y-coordinate of the centre of mass is
$$\eqalign{ & {y_{cm}} = \frac{{{m_1}{y_1} + {m_2}{y_2} + {m_3}{y_3} + {m_4}{y_4} + {m_5}{y_5}}}{{{m_1} + {m_2} + {m_3} + {m_4} + {m_5}}} \cr & \therefore {y_{cm}} = \frac{{6m \times 0 + m \times 0 + m \times a + m \times a + m\left( { - a} \right)}}{{6m + m + m + m + m}} \cr & = \frac{{ma}}{{10m}} \cr & = \frac{a}{{10}} \cr} $$

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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