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JEE Advanced 2009 · System of Particles and Rotational Motion

A sphere is rolling without slipping on a fixed horizontal plane surface. In the figure, AA is the point of contact, BB is the centre of the sphere and CC is its topmost point. Then,

Answers: B. V⃗C−V⃗B=V⃗B−V⃗A\vec{V}_C - \vec{V}_B = \vec{V}_B - \vec{V}_A · C. ∣V⃗C−V⃗A∣=2∣V⃗B−V⃗C∣|\vec{V}_C - \vec{V}_A| = 2|\vec{V}_B - \vec{V}_C|

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Written solution

Step 1: Determine the velocities of points A, B, and C.

For a sphere of radius RR rolling without slipping on a fixed horizontal surface with center velocity V⃗B=vi^\vec{V}_B = v\hat{i} and angular velocity ω=vR\omega = \frac{v}{R} in the clockwise direction:

The point of contact AA is instantaneously at rest.

V⃗A=0\vec{V}_A = 0

The center of the sphere BB moves with velocity vv.

V⃗B=vi^\vec{V}_B = v\hat{i}

The topmost point CC has both translational and rotational velocity components.

V⃗C=V⃗B+ωRi^\vec{V}_C = \vec{V}_B + \omega R\hat{i} V⃗C=vi^+vi^=2vi^\vec{V}_C = v\hat{i} + v\hat{i} = 2v\hat{i}

Step 2: Evaluate Option (1).

Check the relation V⃗C−V⃗A=2(V⃗B−V⃗C)\vec{V}_C - \vec{V}_A = 2(\vec{V}_B - \vec{V}_C).

LHS=V⃗C−V⃗A=2vi^−0=2vi^\text{LHS} = \vec{V}_C - \vec{V}_A = 2v\hat{i} - 0 = 2v\hat{i} RHS=2(V⃗B−V⃗C)=2(vi^−2vi^)=−2vi^\text{RHS} = 2(\vec{V}_B - \vec{V}_C) = 2(v\hat{i} - 2v\hat{i}) = -2v\hat{i}

Since LHS≠RHS\text{LHS} \neq \text{RHS}, Option (1) is incorrect.

Step 3: Evaluate Option (2).

Check the relation V⃗C−V⃗B=V⃗B−V⃗A\vec{V}_C - \vec{V}_B = \vec{V}_B - \vec{V}_A.

LHS=V⃗C−V⃗B=2vi^−vi^=vi^\text{LHS} = \vec{V}_C - \vec{V}_B = 2v\hat{i} - v\hat{i} = v\hat{i} RHS=V⃗B−V⃗A=vi^−0=vi^\text{RHS} = \vec{V}_B - \vec{V}_A = v\hat{i} - 0 = v\hat{i}

Since LHS=RHS\text{LHS} = \text{RHS}, Option (2) is correct.

Step 4: Evaluate Option (3).

Check the relation ∣V⃗C−V⃗A∣=2∣V⃗B−V⃗C∣|\vec{V}_C - \vec{V}_A| = 2|\vec{V}_B - \vec{V}_C|.

LHS=∣V⃗C−V⃗A∣=∣2vi^∣=2v\text{LHS} = |\vec{V}_C - \vec{V}_A| = |2v\hat{i}| = 2v RHS=2∣V⃗B−V⃗C∣=2∣vi^−2vi^∣=2v\text{RHS} = 2|\vec{V}_B - \vec{V}_C| = 2|v\hat{i} - 2v\hat{i}| = 2v

Since LHS=RHS\text{LHS} = \text{RHS}, Option (3) is correct.

Step 5: Evaluate Option (4).

Check the relation ∣V⃗C−V⃗A∣=4∣V⃗B∣|\vec{V}_C - \vec{V}_A| = 4|\vec{V}_B|.

LHS=∣V⃗C−V⃗A∣=2v\text{LHS} = |\vec{V}_C - \vec{V}_A| = 2v RHS=4∣V⃗B∣=4v\text{RHS} = 4|\vec{V}_B| = 4v

Since 2v≠4v2v \neq 4v, Option (4) is incorrect.

Answer: The correct relations are (2) V⃗C−V⃗B=V⃗B−V⃗A\vec{V}_C - \vec{V}_B = \vec{V}_B - \vec{V}_A and (3) ∣V⃗C−V⃗A∣=2∣V⃗B−V⃗C∣|\vec{V}_C - \vec{V}_A| = 2|\vec{V}_B - \vec{V}_C|.

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