Calculating Time for Mass to Travel Distance on a Pulley

[David Mordigal]

Homework Statement

The two blocks in the figure are connected by a massless rope that passes over a pulley. The pulley is 16cm in diameter and has a mass of 1.3kg. As the pulley turns, friction at the axle exerts a torque of magnitude 0.50 N*m.

If the blocks are released from rest, how long does it take the 4.0kg block to reach the floor?
Express your answer using two significant figures.

Homework Equations

τ = Iα
I = (1/2)MR² = (1/2)(1.3kg)(0.08²) = 0.00416 kg*m²

m_pulley = 1.3kg
r_pulley = (16/2)cm = 8cm = 0.08m

The Attempt at a Solution

a. Find the net torque being exerted on the pulley:
Στ = τ_m1 - τ_m2 - τ_ƒ
= F_m1*r*sin(90) - F_m2*r*sin(90) - 0.5 N*m
= (4.0kg)(9.8m/s²)(0.08m) - (2.0kg)(9.8m/s²)(0.08m) - 0.5 N*m
= 3.136 N*m - 1.568 N*m - 0.5 N*m
= 1.068 N*m

b. Find the angular acceleration α of the pulley:
Given τ = Iα, α = τ/I
= (1.068 N*m) / (0.00416 kg*m²)
= 256.73 rad/s² (this seems very large to me)

c. Given the radius of the pulley and α, we can now find linear acceleration:
a = rα = (0.08 m)(256.73 rad/s²) = 20.54 m/s²

d. Use the kinematic equation to find the final velocity after the block travels 1m to the floor:
v² = v₀² + 2ax
v² = 2ax = 2(20.54 m/s²)(1.0 m) = 41.08
v = 6.41 m/s

e. Now substitute v into the definition of velocity and solve for Δt:
v = Δx/Δt
Δt = Δx / v = (1.0 m) / (6.41 m/s)
= 0.16s (rounded to 2 sig figs)

My homework system says this answer is wrong. I am not sure why I keep getting the wrong answer. Can someone help me out here? It is much appreciated.

 

[_N3WTON_]

It looks like your not really on the right track to me. First find three equations, two force equations (for each block) and a torque equation (for the pulley). You should end up with three equations and three unknowns (two of which can be eliminated). At this point its just an algebra problem, so solve for your desired variable and then use one of the kinematic equations to find time.