# Physics rotational motion

• Jamest39
No, it depends on R and U. Do you know how to draw a force diagram, with two forces at right angles, the lengths of the lines representing the magnitudes of the forces? What line represents the resultant? What angle does it make to the other lines?In summary, the homework statement is discussing a sphere that rotates with constant angular acceleration about a fixed axis. A particle within this sphere experiences a tangential acceleration and a centripetal acceleration. The angular velocity is also changing, but considering these vectors as a whole, the direction is constantly changing.

## 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.

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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Jamest39 said:
tangential acceleration = αr
It would be well to consider that as a vector equation. In that view, what varies?

haruspex said:
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.

Jamest39 said:
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?

haruspex said:
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

Jamest39 said:
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.

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

Jamest39 said:
ω in terms of t, so (Δθ)/(Δt)
Use the fact that α is constant.

haruspex said:
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).

Jamest39 said:
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?

haruspex said:
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

Jamest39 said:
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
haruspex said:
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

Jamest39 said:
A 45 degree angle?
No, it depends on R and U. Do you know how to draw a force diagram, with two forces at right angles, the lengths of the lines representing the magnitudes of the forces? What line represents the resultant? What angle does it make to the other lines?