A hollow, thin-walled cylinder and a solid sphere start from rest and roll without slipping down an inclined plane of length 3 m. The cylinder arrives at the bottom of the plane 2.5 s after the sphere. Determine the angle between the inclined plane and the horizontal.

3.83⁰

Explanation:

First step:

The linear acceleration of a rolling object on an inclined plane is given by the following equation:

where

M = mass of the rolling object

R  = radius of the rolling object

θ = is the angle of the inclination of the plane

I= is the moment if inertia of the object about its centre of mass.

Therefore, let

mass of cylinder = m

mass of the sphere = m

the time taken by the sphere to reach the bottom be t₂

Given that:

length of the inclined plane, S = 3 m

Time taken to reach the bottom t = t₂ + 2.7s

acceleration of a body on the plane:

Value for the hollow cylinder =

Value of I/mR² for solid sphere =

a for cylinder = gsinθ/ (1+1)

                      = gsin θ/2

a for sphere = g sin θ/ (1 + 2/3)

                     5gsin θ/3

from the distance equation s = ut +

3 m =

t =

From equation 1, t₁ = t₂ + 2.5

substituting the values of a for cylinder and a for sphere an solving for θ gives

solving for θ gives θ = 3.83⁰

Therefore, the angle between the inclined plane and horizontal = 3.83⁰

0.11 m/s and 0.28 m/s should be correct