Two boxes hang from a solid disk pulley that is free to rotate as the blocks rise/fall.

The left box has a mass m1 = 4.3 kg and the right box has a mass m2 = 2.1 kg. The pulley has mass m3 = 2.5 kg and radius R = 0.15 m. 1)

1) What is the linear acceleration of the left box? (up is the positive direction) m/s2

2) What is the angular acceleration of the pulley? (let counter-clockwise be positive) rad/s2

3) What is the tension in the string between the left mass and the pulley? N

4) What is the tension in the string between the right mass and the pulley? N

5) The boxes accelerate for a time t = 1.51 s. What distance does each box move in the time 1.51 s? m

6) What is the magnitude of the velocity of the boxes after the time 1.51 s? m/s

7) What is the magnitude of the final angular speed of the pulley? rad/s

1) a = 3.07 m/s²

2) α = 2.05 rad/s²

3) T1 = 28.98 N

4) T2 = 27.05 N

5) x = 3.5 meter

6) vf = 4.64 m/s

7) ωf = 30.93 rad/s

Explanation:

Step 1: Data given

Mass of box 1 = 4.3 kg

Mass of box 2 = 2.1 kg

Mass of pulley = 2.5 kg

Radius = 0.15m

1) What is the linear acceleration of the left box?

A force F at the pulley rim gives an accelerating torque (T = F * r) and an angular acceleration dω/dt (rad/s²)

T = F * r = I*dω/dt

Fp = (I/r) * dω/dt

⇒ with I = ½mr²

⇒ with dω/dt = a/r

Fp = (½mr²/r) * a/r

Fn = ½ma = (4.2-1.9) * 9.81 = 22.56 N

Fn - Fp = (m1 + m2) a

Fn - ½ma = (m1+m2) a

Fn = a (m1 + m2 + ½m)

a = Fn / (m1 + m2 + ½m)

a = 22.56 N / (4.2 + 1.9 + ½*2.5)

a = 3.07 m/s²

2) What is the angular acceleration of the pulley?

angular acceleration α = a/r

angular acceleration α = 3.07 / 0.15 = 2.05 rad/s²

3) What is the tension in the string between the left mass and the pulley?

T1 = m1 * (g-a)

T1 = 4.3 * (9.81-3.07)

T1 = 28.98 N

4) What is the tension in the string between the right mass and the pulley?

T2 = m2 (g+a)

T2 = 2.1 * (9.81 + 3.07)

T2 = 27.05 N

5) The boxes accelerate for a time t = 1.51 s. What distance does each box move in the time 1.51 s?

Initial speed v0 = 0 m/s

Distance x = v0*t + ½at²

x = 0*1.51 + ½ * 3.07*1.51²

x = 3.5 meter

6) What is the magnitude of the velocity of the boxes after the time 1.51 s?

vf = the final speed after a movement of 3.5 meter

vf²=v0²+2ax

vf² = 2*a*x

vf² = 2*3.07*3.5

vf² = 21.49

vf = √ (21.49) = 4.64 m/s

7) What is the magnitude of the final angular speed of the pulley? rad/s

final angular speed ωf = vf / r

ωf = 4.64 m/s / 0.15m

ωf = 30.93 rad/s