Question

A 0.20-kg block and a 0.25-kg block are connected to each other by a string draped over a pulley that is a solid disk of inertia 0.50 kg and radius 0.10 m. When released, the 0.25-kg block is 0.35 m off the ground.

What speed does this block have when it hits the ground?

Express your answer with the appropriate units.

Answer 1

let T1 and T2 be the tensions in the string on the 0.2 kg and 0.25 kg side respectievly

then

equations of motion for m1 = 0.2 kg is

T1-m1g = m1a

T1-(0.2*9.8) = (0.2*a)

T1 =1.96 +(0.2*a) ..........(1)

and for m2 = 0.25 kg

m2g-T2 = m2a

(0.25*9.8)-T2 = 0.25*a

T2 = (0.25*9.8) - (0.25*a)

T2 = 2.45 - (0.25*a)...........(2)

and net torques acting on the pulley is Tnet = I*alpha

moment of inertia is I = (0.5*M*R^2) = (0.5*0.5*0.1^2) = 0.0025 kg-m^2

alpha = a/R

and also Tnet = (T2-T1)*R = 0.0025*(a/R)

T2-T1 = 0.0025*a/(0.1^2) = 0.25*a

2.45 - (0.25*a) - 1.96 - (0.2*a) = 0.25*a

accelaration a = 0.7 m/s^2

initial velocity is u = 0 m/s

distance travelled is S = 0.35 m

then using v^2-u^2 = 2*a*S

v^2-0^2 = 2*0.7*0.35

v = 0.7 m/sec