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**gourish****Member**- Registered: 2013-05-28
- Posts: 153

a smooth sphere of radius R is made to translate in a straight line with a constant acceleration a. A particle kept on the top of the sphere is released from there at zero velocity with respect to the sphere. find the speed of the particle with respect to the sphere as a function of the angle θ it slides..

i hope i will get a quick reply i am breaking my head over this problem....

"The man was just too bored so he invented maths for fun"

-some wise guy

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**ShivamS****Member**- Registered: 2011-02-07
- Posts: 3,648

I would solve it but since I am on a phone and you want a quick answer, google the problem. You will find many full solutions.

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**gourish****Member**- Registered: 2013-05-28
- Posts: 153

i found the solutions online but they are hardly clear it' all getting straight to answer i need to know how to get to the solution... people on this forum usually explain why a certain step is taken to find the answer so i wanted an answer on the forum but i can wait for the answer... thanks Shivam...

"The man was just too bored so he invented maths for fun"

-some wise guy

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**ShivamS****Member**- Registered: 2011-02-07
- Posts: 3,648

This is the diagram

Let the sphere move towards left with an acceleration ‘a’ and let us examine the particle’s motion from the sphere, i.e. assuming the sphere at rest.

Let m = mass of the particle

Now, the particle is moving in a circle on the surface of the sphere.

The free body diagram of the particle is

When the particle has slid through an angle θ, let its velocity be ‘v’.

Tangential acceleration = mdv/dt = macosθ + mgsinθ

We know that v = Rdθ/dt

Therefore, mvdv/dt = macosθ(Rdθ/dt) + mgsinθ(Rdθ/dt)

or mvdv = maRcosθdθ + mgRsinθdθ

Integrating both sides;

v^2 /2 = aRsinθ – gRcosθ+ c

Given that the particle starts from rest, i.e. v = 0 at θ = 0

Therefore, c = +gR

Hence, v^2 = 2aRsinθ – 2gRcosθ+ 2gR

or v = [2R(asinθ – gcosθ +g)]^½

*Last edited by ShivamS (2014-05-27 23:19:31)*

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**gourish****Member**- Registered: 2013-05-28
- Posts: 153

thanks again shivam your explanation is much better then that available online....

P.S. if i want to find the the velocity of the particle from the frame of earth then? and can you tell me what is your last diagram for? it looks like a parabolic motion of some body i don't understand it because the particle is supposed to move "on" the surface of sphere

"The man was just too bored so he invented maths for fun"

-some wise guy

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**ShivamS****Member**- Registered: 2011-02-07
- Posts: 3,648

Sorry, that diagram wasn't needed.

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**mathicINDIA****Member**- Registered: 2017-06-03
- Posts: 7

bhai samajhta kya hai be khud ko?

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