Blocks of mass m 1 and m 2 are connected by a massless string that passes over t

Question

Blocks of mass m1 and m2 are connected by a massless string that passes over the pulley in the figure(Figure 1) . The pulley turns on frictionless bearings, and mass m1 slides on a horizontal, frictionless surface. Mass m2 is released while the blocks are at rest.

Part C

Suppose the pulley has mass mp and radius R. Find the acceleration of m1. Verify that your answers agree with part A if you set mp=0.

Part D

Find the tension in the upper portion of the string. Verify that your answers agree with part B if you set mp=0.

Part E

Find the tension in the lower portions of the string. Verify that your answers agree with part B if you set mp=0.

Explanation / Answer

1) Force on m2

m2g - T = m2a

T = m1 a

m2g - m1a = m2a

a = m2g/(m1 + m2)

2) Tension in the string T

T = m1 m2 g/(m1 + m2)

3) Force on the block m2

m2g - Tlower = m2a

Force on the block m1
Tupper = m1a

Force on the pulling
(Tlower - Tupper)R = I alpha = I a/R

I = 1/2 mp * R^2

(Tlower - Tupper)R = 1/2 * mp * R^2 * a/R

(m2g - m2a - m1a) R = 1/2 *mp *a * R

2*m2g - (2m1 + 2m2)a = mp a

a = 2 m2 g /(2m1 + 2m2 + mp)

if we put mp = 0 then a = m2g/(m1 + m2) which is same.

4) Tupper = m1a = 2m1m2g/(2m1 + 2m2 + mp)

When we put mp = 0 then Tupper = m1m2g/(m1 + m2)

5) Tlower = m2g - m2a = m2g - 2m2*m2g/(2m1 + 2m2 + mp) = m2g(2m1 + 2m2 + mp - 2m2)/(2m1 + 2m2 + mp) = (2m1m2g + mp m2 g)/(2m1 + 2m2 + mp)

if we put mp =0 it will be same Tlower = m1m2g/(m1 + m2)

Related Questions

Navigate

• Browse (All)
• Browse B
• Subjects