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UPSEE 2019 Solved Paper

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Question : 8 of 150

Marks: +1, -0

A disc of mass

M

R

is rolling with angular speed

\omega

on a horizontal plane as shown in figure. The magnitude of the angular momentum of the disc about the origin

O

$\frac{1}{2} M R^{2} \omega$
$\frac{3}{2} M R^{2} \omega$
$M R^{2} \omega$
$2 M R^{2} \omega$

Solution:

Here, disc has two types of motion namely translational and rotational. Hence, angular moment of a rigid body in translational cum rotational

=

angular momentum of centre of mass about origin

(O)+

angular momentum of the body about centre of mass.

\Rightarrow L = M R^{2} \omega + I \dot{\omega}

Here,

I = \frac{M R^{2}}{2}

(for disc

)

and

\omega =

angular velocity of the rigid body

= M R^{2} \omega + \left(\frac{M R^{2}}{2}\right) \omega

= \frac{3}{2} M R^{2} \omega

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