Question

The pulley is a uniform cylinder with mass m3 = 0.400 kg and radius R= 4.00 cm, the other two masses are m1 = 2.00 kg and m2 = 1.00 kg, and α = 35.0 degrees. Assume the rope is massless, there is no slipping of the rope on the pulley, there is no friction between m1 and the incline, and the incline position is fixed.

(a)

What is the acceleration of m1 and m2 (both magnitude and direction)? What is the angular acceleration of the pulley? What are the tensions in the rope (both to the left of the pulley and to the right of the pulley?

(b)

If m2 starts from rest 50.0 cm above the ground, where is it when it has speed 0.500 m/s?

Answer 1

Component of gravitational force on m1, along the incline = m1 g sin theta = 2*9.8*sin35 = 11.24 N
Gravtational force on m2 = m2 g = 9.8 N
As m1 g sin theta > m2 g, m1 will accelerate down along theincline and m2 accelerates upward.
Acceleration a = (m1 g siin theta - m2 g)/(m1 + m2 + I/R²)
here I is moment of inertia of pulley = m3 R²/2
Hence a = (m1 g siin theta - m2 g)/(m1 + m2 + m3/2)
a = (11.24 - 9.8) / ( 2 + 1 +0.2) = 0.45 m/sec²
angular acceleration of pulley, alpha = a/R ( as there is no slipping between pulley and string)
alpha = 0.45/0.04 = 11.25 rad/sec²

b) Using v² = u² + 2 a s ( s is displacement )
displacement of m3 , s = 0.5²/2*0.45 = 0.278 m= 27.8 cm
m3 is 50+27.8 cm = 77.8 cm above ground when it's speed is 0.5 m/s