A uniform circular disc placed on a rough horizontal surface is initially given a velocity v0 and angular velocity ω 0 as shown in figure. The disc comes to rest after moving some distance in the direction of motion. Then (v0/r ω 0) is:

Question

A uniform circular disc placed on a rough horizontal surface is initially given a velocity v0 and angular velocity ω 0 as shown in figure. The disc comes to rest after moving some distance in the direction of motion. Then (v0/r ω 0) is: <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-2nsycpp1okxvtnvk.jpg" /> (A) (1/2) (B) 2 (C) (1/3) (D) 3

Tardigrade Admin · Accepted Answer

From conservation of angular momentum about a point on the surface, Initial angular momentum = final angular momentum: Mv0r-(1/2)Mr2ω 0=0 Therefore (v0/r ω 0)=(1/2)

Q. A uniform circular disc placed on a rough horizontal surface is initially given a velocity $v_{0}$ and angular velocity $ω_{0}$ as shown in figure. The disc comes to rest after moving some distance in the direction of motion. Then $\frac{v _{0}}{r ω _{0}}$ is:

$\frac{1}{2}$

$2$

$\frac{1}{3}$

$3$

From conservation of angular momentum about a point on the surface,
Initial angular momentum = final angular momentum:
$M v_{0} r - \frac{1}{2} M r^{2} ω_{0} = 0$
Therefore $\frac{v _{0}}{r ω _{0}} = \frac{1}{2}$

Q. A uniform circular disc placed on a rough horizontal surface is initially given a velocity v0​ and angular velocity ω0​ as shown in figure. The disc comes to rest after moving some distance in the direction of motion. Then rω0​v0​​ is:

21​

2

31​

3

From conservation of angular momentum about a point on the surface, Initial angular momentum = final angular momentum: Mv0​r−21​Mr2ω0​=0 Therefore rω0​v0​​=21​

Q. A uniform circular disc placed on a rough horizontal surface is initially given a velocity $v_{0}$ and angular velocity $ω_{0}$ as shown in figure. The disc comes to rest after moving some distance in the direction of motion. Then $\frac{v _{0}}{r ω _{0}}$ is:

$\frac{1}{2}$

$2$

$\frac{1}{3}$

$3$

From conservation of angular momentum about a point on the surface,
Initial angular momentum = final angular momentum:
$M v_{0} r - \frac{1}{2} M r^{2} ω_{0} = 0$
Therefore $\frac{v _{0}}{r ω _{0}} = \frac{1}{2}$