Question

a block of mass m = 1.50 kg hanging from a thin rope wrapped around a solid, disk-shaped pulley. The pulley has mass M = 7.80 kg, radius R = 0.127 m, and is free to rotate on a frictionless, horizontal axle. If the block is released from rest, determine the magnitude of its downward acceleration.

Answer 1

To find the block's downward acceleration, we apply Newton's second law for translational and rotational motion. By solving the equations, the magnitude of the downward acceleration of the 1.50 kg block is found to be approximately 2.09 m/s^2.

Step-by-step explanation:

To determine the magnitude of the downward acceleration of the block, we need to consider the forces acting on the system and the rotational inertia of the pulley. The gravitational force on the block provides the downward force which is partially countered by the inertia of the pulley.

Let’s denote:
T as the tension in the rope,
g as the acceleration due to gravity (9.81 m/s2),
a as the linear acceleration of the block, and
α as the angular acceleration of the pulley.

Using Newton’s second law for the block and the rotational equivalent for the pulley, we can write two equations:

1. For the block: T = m(g - a)
2. For the pulley: TR = Iα, where I is the moment of inertia of the pulley (I = ½MR2 for a solid disc).

Since the rope does not slip, the linear acceleration a and the angular acceleration α are related by a = αR. Substituting I and then α we get:

T = (½MR2)(a/R)

Combining the equations for T and solving for a:

m(g - a) = ½MR(a/R)
mg = ma + ½Ma^2/R
a(½M/R + m) = mg
a = ½m(g/R)/(½M/R + m)
Now plug in the values:a = ½(1.50 kg)(9.81 m/s2)/(½(7.80 kg)/0.127 m + 1.50 kg)a ≈ 2.09 m/s2

The magnitude of the downward acceleration of the block is approximately 2.09 m/s2.