Physics Quiz 9

[Ether]

Given:

Robot with 4 standard (non-omni non-mec) wheels.

No drivetrain (no gearbox, no chains; just wheels with frictionless bearings)

Total mass of robot *including the wheels* is Mrobot

Total mass of the 4 wheels is Mwheels

Assume that when you push the robot, the wheels rotate (instead of sliding on the floor)

Assume that the mass of the wheels is evenly distributed,
like a solid disc rather than a hoop.

Problem:

Find the "equivalent mass", Me, such that

F = Me*a

...where "a" is the acceleration and "F" is the pushing force.

[sanddrag]

Me = Mrobot + 0.5*Mwheels

I have the solution steps (on 3 post-it notes), but I'd like to see if this is correct first.

[SenorZ]

I'm getting Mwheel/2

[T^2]

Let F_linear be the force that accelerates the robot body, and F_angular be the force that accelerates the rotation of the wheels. F_linear + F_angular = F.

F_linear = (M_robot)(a)

F_angular = tau/r = 0.5(M_wheel)(r)(alpha) = 0.5(M_wheel)(a) where r is wheel radius and alpha is angular acceleration

M_e = (F_linear + F_angular)/a = ((M_robot)(a) + 0.5(M_wheel)(a))/a = M_robot + 0.5(M_wheel)

Not sure if I used M_robot in the equation correctly.

[TRWSHSHLX]

Second sanddrag.

Thank you for posting this quiz so I can review my kinematics and procrastinate for a bit .

[Ether]

That was quick. Nice work guys.

An am-0940 wheel weighs about 0.5 lb. So the extra force due to wheel rotation required to accelerate a 150 lb robot amounts to less than 1%.

[sanddrag]

This was a fun one, because it has been YEARS since I've done something like this. I did conceptual Physics in 2001, AP Physics in 2003, and Physics for three quarters in college in 2005, and then finished my Mechanical Engineering degree in 2010, but I haven't done a whole lot of this type of problem solving since. I had a very rough idea of the concepts going on here, and I recalled a few things from school, but I had to look up the moment of inertia equation for a solid cylinder, and I had to remind myself that Tau = I*alpha.

Initially, I had misread the problem to think that Mwheels was the weight of each wheel, so my answer was 4 times too large, which I quicklyrevised.

Thanks for making me think on a Sunday morning.

[Oblarg]

If anyone's interested, here's a somewhat similar, slightly tricky question I recall from a classical mechanics course:

A bowler releases a bowling ball. Initially, it is sliding at velocity v. Due to friction with the ground, it begins to roll. What is the velocity of the ball once it is rolling?

Hint: Not all origin locations are created equal.

[magnets]

I don't think you've given enough information. I am fairly certain (but may be wrong) that the force of friction on the ball is important, so we'd need to know the mass/friction.

[seg9585]

Originally Posted by Oblarg

If anyone's interested, here's a somewhat similar, slightly tricky question I recall from a classical mechanics course:

A bowler releases a bowling ball. Initially, it is sliding at velocity v. Due to friction with the ground, it begins to roll. What is the velocity of the ball once it is rolling?

Hint: Not all origin locations are created equal.

Should we assume the bowling ball is a solid sphere? Because as a league bowler I'm well aware they are not :-)

[James Kuszmaul]

Originally Posted by Oblarg

If anyone's interested, here's a somewhat similar, slightly tricky question I recall from a classical mechanics course:

A bowler releases a bowling ball. Initially, it is sliding at velocity v. Due to friction with the ground, it begins to roll. What is the velocity of the ball once it is rolling?

Hint: Not all origin locations are created equal.

Well, once the bowling ball is rolling, you will have an angular momentum:
L = I * omega
and a momentum:
p = m * v_f
where v_f = omega * r
This angular momentum must result from some constant force of kinetic friction, f_k acting over a time, t, at the radius r such that:
f_k * r * t = L
This same force of friction will have created an impulse, which slowed down the ball as a whole:
f_k * t = m * v_f - m * v
I * omega / r = m * (v_f - v)
I * v_f = m * (v_f - v)
and so:
v_f = -mv / (I - m)
if we take the bowling ball to actually be a perfect, uniform sphere, then:
v_f = -mv / (.4 * m * r² - m)
so
v_f = -v / (.4 * r² - 1)
Which looks really weird to me, so I have no idea if it is correct.
Edit: Figured out two issues. I forgot a negative sign on the change in translational momentum and I switched where the r should go when converting between angular and linear velocity.
Starting from:
I * omega / r = -m * (v_f - v)
omega = v_f/r
I * v_f / r² = -m * (v_f - v)
v_f = mv / (I / r² + m)
Assuming uniform sphere:
v_f = mv / (0.4 * m * r² / r² + m)
v_f = v / (0.4 + 1)
v_f = 5 * v / 7

[Oblarg]

Originally Posted by seg9585

Should we assume the bowling ball is a solid sphere? Because as a league bowler I'm well aware they are not :-)

Yes, you should assume that.

Originally Posted by James Kuszmaul

v_f = -v / (.4 * r² - 1)
Which looks really weird to me, so I have no idea if it is correct.

Nope, not quite! You're making it a bit harder than it has to be by attempting to calculate from the forces themselves - rather, try to think of a way that you can find the answer simply by analyzing the system in two states only (initial, and final). Placement of the origin is crucial to simplifying the problem.

[seg9585]

Assuming a solid sphere, I get a final velocity v2 = 5/7*v1

Using conservation of momentum:

m*v1*R = I*omega
omega = v2/R (ball rolling v2 = r*omega)

I = I_sphere + I_point mass = 2/5*m*R^2 + m*R^2 = 7/5*m*R^2 (parallel axis theorem)

m*v1*R = 7/5*m*R^2 * v2/R
m, R both cancel

v1 = 7/5*v2 --> v2 = 5/7*v1

[James Kuszmaul]

Originally Posted by Oblarg

Nope, not quite! You're making it a bit harder than it has to be by attempting to calculate from the forces themselves - rather, try to think of a way that you can find the answer simply by analyzing the system in two states only (initial, and final). Placement of the origin is crucial to simplifying the problem.

I'm not entirely sure what you mean by trying "to calculate from the forces themselves" (I only mentioned a force at all so that I could create the equivalency; there may be some convenient equation that already takes that into account). My issues resulted from a couple of silly mental errors in typing out the solution (I shouldn't solve equations in text editors).

[Oblarg]

Originally Posted by James Kuszmaul

I'm not entirely sure what you mean by trying "to calculate from the forces themselves" (I only mentioned a force at all so that I could create the equivalency; there may be some convenient equation that already takes that into account). My issues resulted from a couple of silly mental errors in typing out the solution (I shouldn't solve equations in text editors).

I was trying to get at that if you place your origin on the floor, you can notice that the friction force is pointing towards the origin and thus angular momentum is conserved; you need not not consider any forces in any capacity past that, only the initial and final velocities.

Originally Posted by seg9585

Assuming a solid sphere, I get a final velocity v2 = 5/7*v1

Using conservation of momentum:

m*v1*R = I*omega
omega = v2/R (ball rolling v2 = r*omega)

I = I_sphere + I_point mass = 2/5*m*R^2 + m*R^2 = 7/5*m*R^2 (parallel axis theorem)

m*v1*R = 7/5*m*R^2 * v2/R
m, R both cancel

v1 = 7/5*v2 --> v2 = 5/7*v1

This is correct.