A ball rolling down a slope, find its velocity.

[Milsomonk]

Hi guy's, I've got an interesting problem in my coursework, I have to say I am stumped, so here it is.

A hollow sphere of mass 56 grams, starts at rest and is allowed to roll without slipping 2 metres down a slope at an angle of 25 degrees. Gravity is assumed as 10 ms^2. clculate the velocity after 2 metres.

I=2/3*wr^2 is given as a relevant equation.

I'm not even sure where to start without being given a measurement for r or any dimensions. I've calculated the gravitational potential energy in the hope i can find the proportions of KE(rotational) and KE(linear), But I am at a dead end there.

Any help would be hugely appreciated as I'm baffled.

Cheers

 

[Doc Al]

What might be conserved, since there is no slipping?

Tip: When not given a quantity, such as r, just assume you may not need it. Just solve it symbolically and see what happens.

 

[Milsomonk]

Will it still accelerate at 10 ms^2 straight down? despite the slope? so then we have a time component for the distance.

 

[Doc Al]

Will it still accelerate at 10 ms^2 straight down? despite the slope?

No. 10 m/s^2 would be the acceleration of an object in free fall, not something in contact with a slope.

To solve for the acceleration, if you would like to do that, you'll need to consider all the forces acting and apply Newton's 2nd law for translation and rotation.

But there's an easier way. Answer the question I raised in earlier post.

 

[Milsomonk]

Sorry, I was trying to understand your initial question, (Im not good at this), So the energy is conserved? MGH? I've worked that out but that's where I am stuck.

 

[Doc Al]

Sorry, I was trying to understand your initial question, (Im not good at this), So the energy is conserved? MGH? I've worked that out but that's where I am stuck.

Yes, the energy is conserved.

At the top of the ramp, the energy is all potential. But what about at the bottom? How can you express the total energy of the rolling ball at the bottom?

 

[Milsomonk]

I get that, some will be rotational and some will be linear. Its just a case of working out the proportion of each, and then finding a velocity from that that I am stuck with. Thanks fo the help by the way, I am getting there i think.

 

[Doc Al]

I get that, some will be rotational and some will be linear. Its just a case of working out the proportion of each, and then finding a velocity from that that I am stuck with.

That's exactly it. Hint: The two energies are connected by the fact that the ball rolls without slipping. How do you express that mathematically?

 

[Milsomonk]

PE=MGH so MGH=KE(rotational)+KE(linear). so if there is no slip, does that make the split 50/50?

 

[Tanya Sharma]

PE=MGH so MGH=KE(rotational)+KE(linear). so if there is no slip, does that make the split 50/50?

The relation is correct but they are not in the ratio 1:1 .

Express the relation symbolically in terms of the given parameters like 'm' ,'v', 'R' ...

Use the rolling without slipping constraint.

 

[Milsomonk]

Ok so I think I might be getting somewhere, just one smal hurdle. Heres what I've got.

mgh=1/2(2/3mr^2)(V/r)^2+1/2mv^2

Then I canceled the m from both sides gh=1/2(2/3r^2)(V/r)^2+1/2v^2

Where I'm stuck is how to get rid of r.

 

[Tanya Sharma]

Ok so I think I might be getting somewhere, just one smal hurdle. Heres what I've got.

mgh=1/2(2/3mr^2)(V/r)^2+1/2mv^2

Then I canceled the m from both sides gh=1/2(2/3r^2)(V/r)^2+1/2v^2

Where I'm stuck is how to get rid of r.

There is an r² term in both the numerator and denominator of the first term on the Right Hand Side .

You missed cancelling them

 

[Milsomonk]

AHA silly me, awesome thankyou all for your help I have now done it! only took me three days haha :)