A uniform solid disk of mass m=3.0 kg and radius r=0.20 m rotates about a fixed axis perpendicular to its face with angular frequency 6 rad / s. The magnitude of the angular momentum of the disk when the axis of rotation passes through a point midway between the centre and the rim is

Question

A uniform solid disk of mass m=3.0 kg and radius r=0.20 m rotates about a fixed axis perpendicular to its face with angular frequency 6 rad / s. The magnitude of the angular momentum of the disk when the axis of rotation passes through a point midway between the centre and the rim is (A) 0.72 kg ⋅ m 2 / s (B) 0.54 kg ⋅ m 2 / s (C) 0.36 kg ⋅ m 2 / s (D) 1.08 kg ⋅ m 2 / s

Tardigrade Admin · Accepted Answer

For a point midway between the centre and the rim, we use the parallel-axis theorem to find the moment of inertia about this point. Then,

L = I ω = [\frac{1}{2} M R^{2} + M (\frac{R}{2})^{2}] ω

= \frac{3}{4} (3.00 k g) (0.200 m)^{2} (6.00 r a d / s)

= 0.540 k g \cdot m^{2} / s

$0.72 k g \cdot m^{2} / s$

$0.54 k g \cdot m^{2} / s$

$0.36 k g \cdot m^{2} / s$

$1.08 k g \cdot m^{2} / s$

For a point midway between the centre and the rim, we use the parallel-axis theorem to find the moment of inertia about this point. Then,
$L = I ω = [\frac{1}{2} M R^{2} + M (\frac{R}{2})^{2}] ω$
$= \frac{3}{4} (3.00 k g) (0.200 m)^{2} (6.00 r a d / s)$
$= 0.540 k g \cdot m^{2} / s$

Q. A uniform solid disk of mass m=3.0kg and radius r=0.20m rotates about a fixed axis perpendicular to its face with angular frequency 6rad/s. The magnitude of the angular momentum of the disk when the axis of rotation passes through a point midway between the centre and the rim is

0.72kg⋅m2/s

0.54kg⋅m2/s

0.36kg⋅m2/s

1.08kg⋅m2/s

For a point midway between the centre and the rim, we use the parallel-axis theorem to find the moment of inertia about this point. Then, L=Iω=[21​MR2+M(2R​)2]ω =43​(3.00kg)(0.200m)2(6.00rad/s) =0.540kg⋅m2/s

$0.72 k g \cdot m^{2} / s$

$0.54 k g \cdot m^{2} / s$

$0.36 k g \cdot m^{2} / s$

$1.08 k g \cdot m^{2} / s$

For a point midway between the centre and the rim, we use the parallel-axis theorem to find the moment of inertia about this point. Then,
$L = I ω = [\frac{1}{2} M R^{2} + M (\frac{R}{2})^{2}] ω$
$= \frac{3}{4} (3.00 k g) (0.200 m)^{2} (6.00 r a d / s)$
$= 0.540 k g \cdot m^{2} / s$