A disc shaped body (having a hole as shown in the figure) of mass m = 10 kg and radius R = 9 10 m is performing pure rolling motion on a rough horizontal surface. In the figure point O is geometrica

Question

A disc shaped body (having a hole as shown in the figure) of mass $m = 10 kg$ and radius $R = \frac{10}{9} m$ is performing pure rolling motion on a rough horizontal surface. In the figure point O is geometrical center of the disc and at this instant the centre of the mass C of the disc is at same horizontal level with O. The radius of gyration of the disc about an axis passing through C and perpendicular to the plane of the disc is $\frac{R}{2}$ and at the instant shown the angular velocity of the disc is $ω = \frac{g}{R}$ rad/sec in clockwise sense. g is gravitation acceleration $= 10 m / s^{2}$ . Find angular acceleration $α$ (in $rad / s^{2}$ ) of the disc at this instant.

Answer:

6

Solution:

Considering torque of real and pseudo force in the frame of P,

$i = mg \frac{R}{2} + m ω^{2} R \frac{R}{2} = mg \frac{R}{2} + m \frac{g}{R} R \frac{R}{2} = mgR I_{p} = m (\frac{R}{2})^{2} + m (R^{2} + \frac{R ^{2}}{4}) = \frac{3}{2} mR^{2} ∴ α = \frac{2 mgR}{3 mR ^{2}} = \frac{2 g}{3 R} = 6 rad / sec^{2}$

TestHub · Accepted Answer

Considering torque of real and pseudo force in the frame of P,

$i = mg \frac{R}{2} + m ω^{2} R \frac{R}{2} = mg \frac{R}{2} + m \frac{g}{R} R \frac{R}{2} = mgR I_{p} = m (\frac{R}{2})^{2} + m (R^{2} + \frac{R ^{2}}{4}) = \frac{3}{2} mR^{2} ∴ α = \frac{2 mgR}{3 mR ^{2}} = \frac{2 g}{3 R} = 6 rad / sec^{2}$

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