Question

Three identical metal balls each of radius $$r$$ are placed touching each other on a horizontal surface such that an equilateral triangle is formed with centres of three balls joined. The centre of mass of the system is located at

A. horizontal surface
B. centre of one of the balls
C. line joining the centres of any two balls
D. point of intersection of the medians  
Answer :   point of intersection of the medians
Solution :
The whole mass of the ball will be concentrated at the centre of the ball. All the three balls are identical, i.e., the balls have same mass. On each vertex of equilateral $$\Delta PQR,$$  same mass is kept.
Rotational Motion mcq solution image
Therefore, centre of mass of the triangle is the centre of mass of the system which is point of intersection of the medians of the triangle.

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. $$\frac{{\omega \left( {M + 2m} \right)}}{M}$$
View Answer
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 $$0.7 \,kg$$
View Answer
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. $${\omega _B} = \omega $$
View Answer
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. $$2M{R^2}\omega $$
View Answer

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Rotational Motion

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