A cord is wrapped around the rim of a cylinder of radius 18 cm. When the cord is

Question

A cord is wrapped around the rim of a cylinder of radius 18 cm. When the cord is pulled with a 20 N force, the cylinder rotates with a constant angular acceleration of 1.4 rad/s2. What is the torque due to the pull force If a friction torque of 0.80 m N acts on the wheel, what is the moment of inertia of the wheel At the instant the cylinder has an angular velocity of 24 rad/s, find: The angle that the cylinder has turned through if it started from rest. The centripetal acceleration of a point on the rim. The time it took to accelerate to 24 rad/s from rest.

Explanation / Answer

a) Torque = r * F sin (theta)

theta is angle between r and F

Torque = (18/100) * 20 * sin 90 =3.6 N-m

b) Total Torque (T_tot) = moment of inertia (I) * angular acceleration

I = T_tot/ang. acceleration = (T_pull-T_frictional)/ang acceleration = 2 kg -m^2

c) angular velocity,w = v*sin(theta)/r

sin (theta)= rw/v, now at rest v=rw

sin (theta) = 1

theta =90 degree.

d) centripetal acceleration =w^2 *r = 24^2 * 0.18 = 103.68 rad/s^2

e) acceration = (delta v)/(delta t)

delta t = delta v/ acceleration

delta t = 0.235 sec

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