The velocity of centre of mass of a disc rolling on an inclined plane changes from V to 2 V at any instant of time. If M is the mass of disc then increases in its kinetic energy will be

Question

The velocity of centre of mass of a disc rolling on an inclined plane changes from V to 2 V at any instant of time. If M is the mass of disc then increases in its kinetic energy will be (A) (9 / 4) MV 2 (B) M V2 / 2 (C) MV 2 (D) 3 M V2

Tardigrade Admin · Accepted Answer

Total KE =(1/2) MV 2+(1/2) I ω2=(3/4) MV 2 Δ K E=(3/4) M(4 V2-V2)=(9/4) M V2

Q. The velocity of centre of mass of a disc rolling on an inclined plane changes from $V$ to $2 V$ at any instant of time. If $M$ is the mass of disc then increases in its kinetic energy will be

$(9/4) M V^{2}$

$M V^{2} /2$

MV $^{2}$

$3 M V^{2}$

Total $K E = \frac{1}{2} M V^{2} + \frac{1}{2} I ω^{2} = \frac{3}{4} M V^{2}$
$Δ K E = \frac{3}{4} M (4 V^{2} - V^{2}) = \frac{9}{4} M V^{2}$

Q. The velocity of centre of mass of a disc rolling on an inclined plane changes from V to 2V at any instant of time. If M is the mass of disc then increases in its kinetic energy will be

(9/4)MV2

MV2/2

MV 2

3MV2