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Arlecino [84]
3 years ago
15

A 19 kg solid disk of radius0.44 m is rotated about an

Physics
1 answer:
Vlad1618 [11]3 years ago
8 0

Answer:

0.915 Nm

Explanation:

1 revolution = 2π rad

We can use the following equation of motion to find out the acceleration acting on the disk

\omega^2 - \omega_0^2 = 2\alpha\Delta \theta

where \omega v = 2.5 rad/s is the final angular velocity of the disk, [tex]\omega_0 = 0 rad/s is the initial velocity of the can when it starts from rest, \Delta \theta is the angular distance traveled, \alpha is the angular acceleration of the disk, which we care looking for:

2.5^2 - 0 = 2*\alpha*2\pi

\alpha = \frac{2.5^2}{2*2\pi} \approx 0.5 rad/s^2

The moment of inertia of the solid disk is:

I = \frac{1}{2}mR^2 = \frac{1}{2}19*0.44^2 = 1.8392 kgm^2

where m is the mass and R is the radius of the disk

The net torque applied is

T = \alpha*I = 0.5 * 1.8392 = 0.915 Nm

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