Question

a frictionless pulley has a rope going over it the tension in the rope is 1- n on one side and 12 n on the other side what is the angular acceleration of the pulley?

Answer 1

Answer:

16.67 rad/s

Explanation:

if you read the question, you would notice that some requirements are missing, while searching online, the complete question can be found here:

https://www.chegg.com/homework-help/questions-and-answers/frictionless-pulley-modeled-080-solid-cylinder-030-m-radius-rope-going--tension-rope-10-n--q3499646

and here is the complete question:

"a frictionless pulley which can be modeled as a 0.80 solid cylinder with a 0.30 m radius has a rope going over it. The tension in the rope is 10 N on the right side and 12 N on the left side. What is the angular acceleration of the pulley? "

mass (m) = 0.8 kg

radius (r) = 0.3 m

tension on left side (T1) = 10 N

tension on right side (T2) = 12 N

angular acceleration = torque / moment of inertia

where

• torque = (T2-T1) x r

        torque = (12-10) x 0.3 = 0.6 Nm

• moment of inertia = 0.5m$r^{2}$

        moment of inertia = 0.5 x 0.8 x $0.3^{2}$ = 0.036 kg

therefore

angular acceleration = 0.6/0.036 = 16.67 rad/s