Question

Consider a pulley of mass mp and radius R that has a moment of inertia 1/2mpR2. The pulley is free to rotate about a frictionless pivot at its center. A massless string is wound around the pulley and the other end of the rope is attached to a block of mass m that is initially held at rest on frictionless inclined plane that is inclined at an angle β with respect to the horizontal. The downward acceleration of gravity is g. The block is released from rest .

How long does it take the block to move a distance d down the inclined plane?
Write your answer using some or all of the following: R, m, g, d, mp,

Answer 1

a =
`(m)/(m+ (1)/(2) m_p) \ g \ sin \beta` , t =
`\sqrt{ (2d)/(a) }`

Step-by-step explanation:

To solve this exercise we must use Newton's second law

For the block

let's set a reference system with the x axis parallel to the plane

X axis

Wₓ - T = m a

Y axis

N- W_y = 0

N = W_y

for pulley

∑τ = I α

T R = (½ m_p R²) α

let's use trigonometry for the weight components

sin β = Wₓ / W

cos β = W_y / W

Wx = W sin β

angular and linear variables are related

a = α R

α = a / R

we substitute and group our equations

W sin β - T = m a

T R = ½ m_p R² (a / R)

W sin β - T = m a

T = ½ m_p a

we solve the system of equations

W sin β = (m + ½ m_p) a

a =
`(m)/(m+ (1)/(2) m_p) \ g \ sin \beta`

let's find the time to travel the distance (d) through the block

x = v₀ t + ½ a t²

d = 0 + ½ a t²

t =
`\sqrt{ (2d)/(a) }`