Answer:
3.83⁰
Explanation:
First step:
The linear acceleration of a rolling object on an inclined plane is given by the following equation:
[tex]a = \frac{gsin\theta }{1+ (I/MR^{2}) }[/tex]
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:
[tex]a = \frac{gsin\theta }{1+(I/(mR^{2}) }[/tex]
Value for the hollow cylinder = [tex]\frac{mR^{2} }{mR^{2} } = 1[/tex]
Value of I/mR² for solid sphere = [tex]\frac{2mR^{2} }{3} = 2/3[/tex]
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 + [tex]\frac{1}{2} at^{2}[/tex]
3 m = [tex]0.5at^{2}[/tex]
t = [tex]\sqrt{\frac{6}{a} }[/tex]
From equation 1, t₁ = t₂ + 2.5
[tex]\sqrt{\frac{6}{acylider } } = \sqrt{\frac{10}{a_{sphere} } } + 2.5[/tex]
substituting the values of a for cylinder and a for sphere an solving for θ gives
[tex]\sqrt{\frac{20}{9.81 sin\theta } } = \sqrt{\frac{6}{9.81sin\theta } } + 2.5[/tex]
solving for θ gives θ = 3.83⁰
Therefore, the angle between the inclined plane and horizontal = 3.83⁰
