Archived Rotational Motion -- a hollow ball rolls down a ramp

[Homerr]

Homework Statement

Starting from rest, ahollow ball rolls down a ramp inclined at angle θ to the horizontal. Find an expression for its speed after it's gone a distance d along the incline.

Homework Equations

I for sphere = 2/5*M*R^2

The Attempt at a Solution

I know we have to use v^2=u^2+2ax
and we have to find the angular acceleration however I am not sure how to relate both of them.

 

[Crush1986]

First thing you'll want to note is that the ball is hollow. So 2/5mR^2 is the wrong moment of inertia.

Second I think you should think about conservation of energy.

I'm not a tutor or answer person for the site but I think this should set you on the right track.

 

[Adithyan]

Please mention if the ball does pure rolling or not.

 

[velo city]

The moment of inertia for a hollow sphere is 2/3*M*r^2. not 2/5*M*r^2 that's for a solid sphere.

 

[proton007007]

Please mention if the ball does pure rolling or not.

The word "rolls" implies pure rolling unless otherwise mentioned

 

[hanoon]

Consider the ball starting at rest at a height yi . It’s initial energy is entirely potential energy, so Ei = Mgyi .
As the ball rolls down the incline, to the right, it picks up speed and loses potential energy.
Since the ball is rolling it has both translational and rotational kinetic energy.
So, the final energy is

Ef = KE of trans + KE of rot + M g yf = 1 /2 M v^2 + 1 /2 I ω^2 + M g yf

, where v is the translational speed, I is the moment of inertia, and ω is the rotational speed.
Conservation of energy says that

Mg (yi − yf ) = 1 /2 M v^2 + 1 /2 Iω^2

But, the change in height H = yi − yf .

Furthermore, from the geometry, H = d sin θ. Thus, we find

M g d sin θ = 1 /2 M v^2 + 1/ 2 Iω^2 .

Now, the condition that the ball roll without slipping means that
v = Rω,
where R is the radius of the ball. Thus, we have

M g d sin θ = 1 /2 M v^2 + (1/2) I ( v^2 / R^2) = M /2 (1 + I/ M R^2 ) (v ^2 ).

Now, we are told that the moment of inertia is I = 2 /3 MR^2 ,

and so

M g d sin θ = M /2 (1 + 2 /3) v ^2 = (5 M /6) v ^2 .

Canceling off the mass from both sides and solving for the speed gives

v^2 = (6 /5) g d sin θ

this the answer

 

[physics tutor]

in rolling motions there are two types of kinetic term, translational motion and rotational one. let the ball start from rest at height of $h$ from the base and rolls a distance of $d$ along the incline. Using the conservation of Mechanical energy, we get
$$E_i=E_f \Rightarrow m g h_i=\frac 1 2 I \omega^{2}+\frac 1 2 m V_{CM}^{2}$$
Recall that translational velocity or center of mass velocity is related to the angular velocity by $$V_{CM}=\omega r$$ where $r$ is the distance between axis of rotation and point of impact with the ground which in this case is the radius of the ball. therefore
$$mgd\,\sin \theta=\frac 1 2 I \left(\frac {V_{CM}}{r}\right)^{2}+\frac 1 2 m V_{CM}^2 $$
which gives the velocity of the hollow ball (or translational velocity) at the incline as
$$V_{CM}=\sqrt {\frac {10} 7 gh}=\sqrt {\frac {10} 7 g (d\,\sin \theta)}$$
For more solved problems see,
https://physexams.com/exam/Rotational_Motions/15
https://physexams.com/exam/work_and_energy/3