A 28.7 kg block (m1) is on a horizontal surface, connected to a 7.0- block (m2)

Question

A 28.7 kg block (m1) is on a horizontal surface, connected to a 7.0- block (m2) by a massless string as shown in the Figure. The frictionless pulley has a radius R=0.069 m and a moment of inertia of I= 0.110 kg.m^2. A force, F=224.1 N, acts on m1 at an angle of 28.3 degrees. There is no friction between m1 and the surface. Assume the blocks start from rest and there is no slipping between the pulley and rope.

1. What is the net torque on the pulley???

2. How fast will the pulley be rotating after m1 has been pulled 75 cm????

Please show work so I can follow along. THANK YOU!

Explanation / Answer

Given that

A mass of block (m1) = 28.7 kg

Another block is attached to the string is (m2) = 7.0kg

The frictionless pulley has a radius (R) =0.069 m

The moment of inertia of (I)= 0.110 kg.m^2.

The force applied is , F=224.1 N, acts on m1 at an angle of (theta) = 28.3 degrees

The net torque acting on the pulley is

FApp -m2g =I*alpha

Consier the x-component of force applied is

FAppCostheta*r -m2gr =I*alpha

Then alpha =FAppCostheta*r -m2gr/I =(224.1)cos(28.3)*0.069-(7kg)(9.81m/s2)(0.069)/ 0.110 kg.m^2.=(13.614-4.738)/ 0.110 kg.m^2.=80.688rad/s2

Then the tanbential acceleration is given by

at =alpha*r =(80.688rad/s2)(0.069m)=5.5676m/s2

Then net torque on the pulley is given by

t =I*alpha =(0.110 kg.m^2)(80.688rad/s2)=8.875N.m

2)

The speed of the pulley rotating after m1 has been pulled to a distance of 75cm=0.75m

v2 =2as

v =Sqrt(2as) =Sqrt(2*5.5676*0.75m) =2.889m/s

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