A string is wrapped around a uniform disk of mass M = 1.3 kg and radius R = 0.11

ID: 1556680 • Letter: A

Question

A string is wrapped around a uniform disk of mass

M = 1.3 kg

and radius

R = 0.11 m

(see figure below). Attached to the disk are four low-mass rods of radius

b = 0.15 m,

each with a small mass

m = 0.6 kg

at the end. The device is initially at rest on a nearly frictionless surface. Then you pull the string with a constant force

F = 18 N

for a time of 0.2 s. Now what is the angular speed of the apparatus?
?????? rad/s

Also calculate numerically the angle through which the apparatus turns, in radians and degrees.

radians = ???? rad degrees = ????? ° m. m. M, R

Moemnt of inertia of thr disk = 0.5 MR^2

Idisk = 0.5 * 1.3 * 0.11^2 = 0.007865 kgm^2

MOment of Inertia due to other four masses = 4 * mb^2

I2 = 4 * 0.6 * 0.15^2

I2 = 0.054 kgm^2

Total MOI = Mdisk + M or Four masses

Itot = 0.007865 + 0.054 = 0.061865 kgm^2

torque T = r x F = I A

A is alpha = angular accelration

I is total MOI

A = 0.11* 18/(0.061865)

A = 32 rad/s^2

now use Wf = Wi + A t

at time t = 0 , Wi = 0

at time t = 0.2 s, Wf = 0 + (32* 0.2)

Wf = 6.4 rad/s is the angular speed

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So delta theta = 0.5 A t^2

deltha theta = 0.5 * 32 * 0.2^2

delta theta = 0.64 rad

delta theta = 36.67 deg

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