Someone verify my answer to this question about rotation?

[24forChromium]

Homework Statement

A lever with one end on a pivot point, a point mass on the lever's shaft with a certain arm length (d), it is also the only relevant mass body (m). A torque (Tau_app.) is applied to the lever, an angular acceleration (alpha) transpired. Find the relationship between the force received by the point mass and the other variables.

Homework Equations

alpha = Tau / mass (...)
F_received = -F_{point on shaft} (as the shaft moves, a force is conferred on the point mass, the point mass in return exerts a force on the shaft in the opposite direction)

The Attempt at a Solution

ΣTau = m*alpha
Tau_app. - Tau_resistance = m*alpha
Tau_app. - F_{point on shaft}*d= m*alpha

F_received = (mass*alpha-Tau_app.)/d

 

[Nathanael]

The units of Tau/mass are not units of angular acceleration.

 

[24forChromium]

The units of Tau/mass are not units of angular acceleration.

What is it then? (if it is anything) and what should be the unit of angular acceleration?

 

[Nathanael]

The unit of angular acceleration is (time)^-2

What is the relationship between torque and angular acceleration? It's analogous to F=ma

 

[24forChromium]

The unit of angular acceleration is (time)^-2

What is the relationship between torque and angular acceleration? It's analogous to F=ma

Sorry I was stupid, Tau = MomentOfInertia * angular acceleration

 

[Nathanael]

F_received = -F_{point on shaft} (as the shaft moves, a force is conferred on the point mass, the point mass in return exerts a force on the shaft in the opposite direction)

This is not really a useful equation. You have no way of finding the force of the point mass on the shaft.

Try to find the (linear) acceleration of the point mass so you can use F=ma

 

[24forChromium]

This is not really a useful equation. You have no way of finding the force of the point mass on the shaft.

Try to find the (linear) acceleration of the point mass so you can use F=ma

Why can't I find the force exerted by the point mass on the shaft? What I was thinking is, at a specific moment in time, if there is an angular acceleration, then there must be a net torque, the net torque is made up of components, the applied torque and the resisting torque, the resisting torque, in my opinion, can only be the result of the point mass pushing back against the shaft, thus, by finding the net torque, the applied torque, the arm length to the torque by the point mass, one can find the force of the point mass on the shaft, it is what I am trying to find.

But I guess the angular acceleration of the shaft can be "translated" into the linear acceleration of the point, say it's 90degrees/second^2, and the point mass is 10m away from the axis, since alpha=acceleration/radius, acceleration = alpha*radius = 90°/s² * 10m= 1.59m/s^2
Once we get the linear acceleration, just use the mass to deduce the received force.

 

[Nathanael]

Why can't I find the force exerted by the point mass on the shaft? What I was thinking is, at a specific moment in time, if there is an angular acceleration, then there must be a net torque, the net torque is made up of components, the applied torque and the resisting torque, the resisting torque, in my opinion, can only be the result of the point mass pushing back against the shaft, thus, by finding the net torque, the applied torque, the arm length to the torque by the point mass, one can find the force of the point mass on the shaft, it is what I am trying to find.

The point mass pushes on the shaft and the shaft pushes on the point mass. Both of these (equal and opposite) forces produce an equal and opposite torque which cancels out. These are called "internal forces" because both of forces (the "action and reaction" forces) act within the system, so their effect cancels out and can be ignored. The net torque is due only to external forces.

acceleration = alpha*radius = 90°/s² * 10m

It is important that you measure alpha in radians per second per second (not degrees per second^2) before you multiply it by the radius. This formula (alpha*radius=acceleration) is only true when alpha is measured in radians per second^2

The reason is because the length of an arc on a circle is R*θ if θ is measured in radians (that's essentially the definition of radians and it's why it's such a useful and natural unit.)
(θ is the angle subtended by the arc we are measuring and R is the radius of the circle)

 

[24forChromium]

The point mass pushes on the shaft and the shaft pushes on the point mass. Both of these (equal and opposite) forces produce an equal and opposite torque which cancels out. These are called "internal forces" because both of forces (the "action and reaction" forces) act within the system, so their effect cancels out and can be ignored. The net torque is due only to external forces.It is important that you measure alpha in radians per second per second (not degrees per second^2) before you multiply it by the radius. This formula (alpha*radius=acceleration) is only true when alpha is measured in radians per second^2

The reason is because the length of an arc on a circle is R*θ if θ is measured in radians (that's essentially the definition of radians and it's why it's such a useful and natural unit.)
(θ is the angle subtended by the arc we are measuring and R is the radius of the circle)

Okay, I did that, and I got this:
Force on point mass = 1.59m/s^2

Now, is it correct to say that:
Force on point mass = (Moment of Inertia * angular acceleration - torque_applied) / arm length of point mass
?

 

[Nathanael]

Okay, I did that, and I got this:
Force on point mass = 1.59m/s^2

Units should be kg*m/s^2 but that's probably just a typo.
I can't tell you if this is correct because the problem statement in your OP did not have any numbers in it.

Now, is it correct to say that:
Force on point mass = (Moment of Inertia * angular acceleration - torque_applied) / arm length of point mass
?

Moment of inertia * angular acceleration equals the applied torque, so your equation will always be zero.