Question

Imagine a frictionless pulley in the shape of a solid cylinder of unknown mass M and radius 0.165 m that is used to draw water from a well. A bucket of mass 1.5 kg is attached to a cord wrapped around the cylinder. The bucket starts from rest at the top of the well and falls for 2.5 s before hitting the water at a distance of 6 m below the top of the well. Neglect the mass of the cord.
Required:
(A) What is the linear acceleration of the falling bucket?
Hint: remember that h=(1/2)*a*t^2? a=2h/t^2
(B) What is the angular acceleration of the rotating pulley?
(C) What is the tension in the cord?
(D) What is the value of the torque that is applied to the pulley by the bucket hanging on the cord? Use the tension from the previous question.

Answer 1

Answer:

a) 1.92 m/s2

b) 11.64 rad/s2

c) 14.715 N

d) 2.43 Nm

Explanation:

a) Using the following equation of motion we can solve for the linear acceleration a:

$h = at^2/2$

where h = 6m is the vertical falling distance, and t = 2.5s is the time of falling

$6 = a2.5^2/2$

$a = 12/2.5^2 = 1.92 m/s^2$

b) The angular acceleration of the rotating pulley can be calculated using its radius r = 0.165 m

$\alpha = a/r = 1.92 / 0.165 = 11.64 rad/s^2$

c) As the tension of the cord is caused by the gravity acting on the falling bucket. Let g = 9.81m/s2, the weight of the bucket and tension of the cord would be

F = W = mg = 1.5*9.81 = 14.715 N

d) The torque generated by the tension would be the product of the tension itself and the distance to the pivot point, aka center of the pulley

T =Fr = 14.715 * 0.165 = 2.43 Nm

Answer 2

Answer:

a) a = 1.92 m/s²

b) $\alpha =7.7rad/s$

c) T = 12 N

d) τ = 3 Nm

Explanation:

a) the linear acceleration is

$d=vt+\frac{1}{2} at^{2} \\6=\frac{1}{2}at^{2} \\a=\frac{2*6}{2.5^{2} } =1.92m/s^{2}$

b) the angular acceleration is

$\alpha =\frac{a}{r} =\frac{1.92}{0.25} =7.7rad/s$

c) For the hanging mass the tension is

$mg-T=ma\\T=1.5(9.8-1.92)=12N$

d) the torque is

$t=tension*radius=12*0.25=3Nm$