What Is the Minimum Velocity Needed for a Wheel to Climb a Step?

[TyloBabe]

Homework Statement

A solid wheel of mass M and radius R rolls without slipping
at a constant velocity v until it collides inelastically with a step of height h < R.
Assume that there is no slipping at the point of impact. What is the minimum
velocity v in terms of h and R needed for the wheel to climb the step?

Homework Equations

Not sure. Perhaps conservation of Angular Momentum. We can not use energy because the collision is inelastic.

The Attempt at a Solution

I really have no clue. I know that Angular Momentum is conserved because the only real external force acting on the system is when the disc encounters the corner of the step, but that force always acts radially and so there is no torque.

I can only think that maybe the angular momentum about the discs radius, as well as about the point h, is conserved throughout the entire collision. Where do I begin?

 

[tiny-tim]

Hi TyloBabe!

Perhaps conservation of Angular Momentum. We can not use energy because the collision is inelastic.

Yes and no …

only the collision is inelastic, so you will need to use conservation of https://www.physicsforums.com/library.php?do=view_item&itemid=313" to find the angular velocity immediately after the collision …

but from then on you can assume that energy is conserved.

I really have no clue. I know that Angular Momentum is conserved because the only real external force acting on the system is when the disc encounters the corner of the step, but that force always acts radially and so there is no torque.

No, there is a torque, because the friction is tangential to the disc.

Since you don't know how much the friction is, the place to take moments about is … ?

 

[TyloBabe]

What I was thinking was saying that the initial angular momentum about the center of the disc, would be equal to the angular momentum of the disc about the point of contact on the step. Would that be correct?

I see what you mean about the torque. The normal force on the disc at the step acts radially, but the frictional force from the step acts tangentially in the opposite direction of motion.

Then if I can figure out what that velocity is just after the collision, and then use conservation of energy, i would just have to say that the rotational energy about the step would be equal to the rotational, translational, and gravitational energy just after the disc has made it over the step. Correct me if I'm mistaken, please.

 

[tiny-tim]

Hi TyloBabe!

What I was thinking was saying that the initial angular momentum about the center of the disc, would be equal to the angular momentum of the disc about the point of contact on the step. Would that be correct?

No, you can't change the point about which you measure the https://www.physicsforums.com/library.php?do=view_item&itemid=313"

Choose a point and stick to it!

In this case (btw, you didn't answer my question) that point would be the edge of the step … because there's no unknown https://www.physicsforums.com/library.php?do=view_item&itemid=175" about the edge (the friction has no torque there)

(Don't forget to include the initial translational angular momentum!)

I see what you mean about the torque. The normal force on the disc at the step acts radially, but the frictional force from the step acts tangentially in the opposite direction of motion.

Then if I can figure out what that velocity is just after the collision, and then use conservation of energy, i would just have to say that the rotational energy about the step would be equal to the rotational, translational, and gravitational energy just after the disc has made it over the step. Correct me if I'm mistaken, please.

Sort of … don't forget that you only need the final angular velocity to be zero!