What force is needed to lift a wheel over a bump with given mass and radius?

AI Thread Summary
To determine the force needed to lift a 2kg wheel with radius R over a step of height h (where R is twice h), torque and moment of inertia are key concepts. The pivot point for calculating torque is at the corner of the step, and the applied force is horizontal. Both the applied force and the force of gravity create torque about this pivot. The applied force must at least balance the torque created by gravity to successfully lift the wheel. Focusing solely on torque simplifies the problem, as the goal is to find the minimum force required to lift the wheel over the bump.
Julia123
Messages
3
Reaction score
0

Homework Statement


The question is as follows:
How much force should I apply if i want to have a wheel of mass 2kg and radius R to pass over a step with height h, where R is twice as big as h.
I attached a picture of the problem for reference.
wheel over bump.png

Homework Equations


I am not sure but I think I should be using torque and possibly moment of Inertia to solve this problem.

The Attempt at a Solution


I don't know how to approach the problem and would really appreciate a nudge in the right direction. Thank you in advance for any and all help!
 
Physics news on Phys.org
Here's a nudge: Torque is the way to go. What will you take as your pivot point for calculating torque?
 
The pivot point must be the corner of the step right? But what do I do from there? How can I know that the force is perpendicular if the wheel is spinning over the bump and the angle is changing?
 
Julia123 said:
The pivot point must be the corner of the step right?
That works.

Julia123 said:
How can I know that the force is perpendicular if the wheel is spinning over the bump and the angle is changing?
The applied force is horizontal, per the diagram. What other forces act to create a torque about the pivot?
 
the force of gravity also works to create a torque about the pivot. The moment of inertia works against the angular acceleration of the wheel. Do I have to account for both of these things in the torque equation?
 
Julia123 said:
the force of gravity also works to create a torque about the pivot.
Good. In what direction is the torque due to gravity? The applied force must at least balance that torque in order to lift up the wheel.

Julia123 said:
The moment of inertia works against the angular acceleration of the wheel. Do I have to account for both of these things in the torque equation?
No, forget about angular acceleration. Just consider torque. You are looking for the minimum applied force to just barely lift the thing over the bump.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top