# Physics rotational motion

## Homework Statement

A solid sphere, starting from rest, rotates with constant angular acceleration α about a fixed axis passing through its center. Consider a particle within this sphere that lies at a perpendicular distance r from this axis. If, at some instant, the linear acceleration of this particle makes an angle of 50° with respect to the direction of its tangential acceleration, what is the total angle that the sphere has turned through up until that instant?

## Homework Equations

tangential acceleration = αr
centripetal acceleration = ω^2r

## The Attempt at a Solution

Since the angular acceleration is constant, the tangential acceleration would also remain constant and I don't see how the angle could be changing.

Simon Bridge
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Jamest39 said:
Since the angular acceleration is constant, the tangential acceleration would also remain constant and I don't see how the angle could be changing.
Definition of acceleration.
Note: angular acceleration is constant but not zero... who says the centripetal acceleration is constant?
You are given the final angle... to make sense of the problem statement you need to make a decision about the initial angle and maybe other things too. You'll need to look through your notes for the right context.

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haruspex
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Gold Member
tangential acceleration = αr
It would be well to consider that as a vector equation. In that view, what varies?

It would be well to consider that as a vector equation. In that view, what varies?
The angular velocity would be changing. But considering those as vectors, the direction would be constantly changing.

haruspex
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Gold Member
considering those as vectors, the direction would be constantly changing.
Exactly so. Therefore the tangential acceleration is changing (in direction).
Starting from rest at t=0, what is its tangential acceleration at time t, and what is its radial acceleration at that time?

Exactly so. Therefore the tangential acceleration is changing (in direction).
Starting from rest at t=0, what is its tangential acceleration at time t, and what is its radial acceleration at that time?
Since its starting from rest, it would be zero at t=0. Then after that, the tangential acceleration is given by rα or (Δω)/(Δt) and the centripetal acceleration is given by (ω^2)r

haruspex
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Gold Member
Since its starting from rest, it would be zero at t=0. Then after that, the tangential acceleration is given by rα or (Δω)/(Δt) and the centripetal acceleration is given by (ω^2)r
Yes, but express ω in terms of t.

Yes, but express ω in terms of t.
ω in terms of t, so (Δθ)/(Δt)

haruspex
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Gold Member
ω in terms of t, so (Δθ)/(Δt)
Use the fact that α is constant.

Use the fact that α is constant.
right, so that allows us to use the equation ω=ω(initial) + αt, and since its staring from rest, the ω(initial is zero).

haruspex
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Gold Member
right, so that allows us to use the equation ω=ω(initial) + αt, and since its staring from rest, the ω(initial is zero).
Right. So you know the tangential acceleration at time t and the radial acceleration at time t. So in what direction is the overall linear acceleration at time t?

Right. So you know the tangential acceleration at time t and the radial acceleration at time t. So in what direction is the overall linear acceleration at time t?
In the direction of the rotation, between the tangential and centripetal acceleration vectors

haruspex
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Gold Member
In the direction of the rotation, between the tangential and centripetal acceleration vectors
You can write it as an explicit formula, in terms of r, t and alpha.
It might help if I ask you this first: if there is a vertically upward force U acting on a body and a force R acting horizontally to the right, at what angle to the horizontal is the net force?

• Jamest39
You can write it as an explicit formula, in terms of r, t and alpha.
It might help if I ask you this first: if there is a vertically upward force U acting on a body and a force R acting horizontally to the right, at what angle to the horizontal is the net force?
A 45 degree angle?

i have the same problem and have no clue how to start

haruspex