Question

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

Answer 1

Answer:

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