A 111 g Frisbee is 25 c m in diameter and has half its mass spread uniformly in

Question

A 111g Frisbee is 25cm in diameter and has half its mass spread uniformly in the disk and the other half concentrated in the rim.

A) What is the Frisbee's rotational inertia?

Explanation / Answer

a). rotational inertia is moment of inertia. moment of inertia of disk part = M1*R^2/2 => I1 = [(111*10^-3/2)*(0.25/2)^2]/2 => I1 = 4.336*10^-4 kg-m^2 moment of inertia of rim part = M2*R^2 => I2 = (111*10^-3/2)*(0.25/2)^2 => I2 = 8.672* 10^-4 kg-m^2 total rotational inertia = I1 + I2 = 1.3 *10^-3 kg-m^2 b). rev = 0.25 => theta rotated = 0.25*2*pi= 1.57 radian w = 440 rpm = 440*2*pi/60 rad/s = 46.05 rad/s w^2 - w0^2 = 2*alpha*theta => (46.05)^2 - 0 = 2*alpha*1.57 => alpha= 675.35 rad/s^2 torque = moment of inertia*angular acceleration => T = I*alpha => T = 1.3 *10^-3* 675.35 => T = 0.878 N-m

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