Question

A very thin circular hoop of mass(m) and radius(r) rolls without slipping down a ramp inclined at an angle(theta) with the horizontal, as shown in the figure.What is the acceleration(a) of the center of the hoop? Express your answer in terms of some or all of the variablesm,r, theta, and the magnitude of the acceleration due to gravity(g).

Answer 1

Weight of hoop = mg

Angle of inclination = theta = O

Component of weight parallel to plane = mg sinO

The acceleration of the center of the hoop = a

Force of friction = f

Torque due to friction = t =rf

But torque = t =Ialpha =Ia/r

rf = Ia/r

f =Ia/r^2

ma = mgsinO - f

ma = mgsinO - Ia/r^2

ma + Ia/r^2 =mg sinO

divide by 'm'

a [1+(I/mr^2)] =g sin O

a = gsin O / [1+(I/mr^2)]

Moment of inertia of hoop = I = mr^2

a = gsin O / [1+(mr^2/mr^2)]

a =g sinO/[1+1]

a =0.5 g sinO

Answer 2

A very thin circular hoop of mass m and radius r rolls without slipping down a ramp inclined at an angle theta with the horizontal,

Weight of hoop = mg

Angle of inclination = theta = O

Component of weight parallel to plane = mg sinO

The acceleration of the center of the hoop = a

Force of friction = f

Torque due to friction = t =rf

But torque = t =Ialpha =Ia/r

rf = Ia/r

f =Ia/r^2

ma = mgsinO - f

ma = mgsinO - Ia/r^2

ma + Ia/r^2 =mg sinO

divide by 'm'

a [1+(I/mr^2)] =g sin O

a = gsin O / [1+(I/mr^2)]

Moment of inertia of hoop = I = mr^2

a = gsin O / [1+(mr^2/mr^2)]

a =g sinO/[1+1]

a =0.5 g sinO

Answer 3

Weight of hoop = mg

Angle of inclination = theta = O

Component of weight parallel to plane = mg sinO

The acceleration of the center of the hoop = a

Force of friction = f

Torque due to friction = t =rf

But torque = t =Ialpha =Ia/r

rf = Ia/r

f =Ia/r^2

ma = mgsinO - f

ma = mgsinO - Ia/r^2

ma + Ia/r^2 =mg sinO

divide by 'm'

a [1+(I/mr^2)] =g sin O

a = gsin O / [1+(I/mr^2)]

Moment of inertia of hoop = I = mr^2

a = gsin O / [1+(mr^2/mr^2)]

a =g sinO/[1+1]

a =0.5 g sinO