Question

A thin spherical shell rolls down an incline without slipping. If the linear acceleration of the center of mass of the shell is 0.18g, what is the angle the incline makes with the horizontal?

Answer 1

`17.46^(\circ)`

Step-by-step explanation:

`a_(cm)=\frac {gsin\theta}{1+\frac {I}{MR^(2)}}` but for sphere
`I=\frac {2MR^(2)}{3}` where
`a_(cm)` is linear acceleration, g is acceleration due to gravity whose value is taken as
`9.81 m/s^(2)`,
`\theta` is the angle of inclination, M is mass of object and R is radius, I is rotational inertia.

Substituting 0.18 g for
`a_(cm)` we obtain

`0.18g=\frac {gsin\theta}{1+\frac {I}{MR^(2)}}=\frac {gsin\theta}{1+\frac {\frac {2MR^(2)}{3} }{MR^(2)}}`

`0.18g=\frac {gsin \theta}{\frac {5}{3}}`

`0.18*9.81*\frac {5}{3}=9.81 sin\theta`

`\theta=sin^(-1)(0.304692654)=17.45760312^(\circ)\approx 17.46^(\circ)`