Find the Minimum Velocity Needed for Ball to Go Up Step - Hint Included!

AI Thread Summary
To determine the minimum velocity needed for a ball to ascend a step without slipping, consider the momentum equilibrium at the contact point between the ball and the step. The condition of no slipping implies that the velocity of the contact point remains zero during motion. Analyzing the collision phase reveals that angular momentum is conserved, which helps calculate the angular velocity after the collision. In the post-collision phase, mechanical energy conservation applies, as only gravity does work on the system. Applying these principles will lead to the required minimum velocity for the ball to successfully navigate the step.
Feynmanfan
Messages
128
Reaction score
0
What is the minimum velocity the ball should have in order to go up the step, considering it doesn't slip. Here's a picture.

I would be grateful if you could give me a hint.
 

Attachments

Physics news on Phys.org
is it that i didn't explain the problem correctly?

Please help me!
 
Your picture is very diffuse. Anyway, If I were you, I will take the momentum (torque) equilibrium at the point of contact between the ball and step. At that point

\sum F_i r_i >0 in order to enhance the turning of the ball around the wedge.

Also, you should try with energy conservation (kinetic--->potential) if you want a gross estimation.
 
You need to break this up a bit:
You have been told that the ball will NOT slip at the corner; evidently this must mean that the velocity of the contact point of the ball is ZERO, throughout the motion (after contact is has been made)

A) The collision phase:
The angular momentum of the ball with respect to the corner must be conserved throughout the collision (why is that).
This gives you the resultant angular velocity just after the collision, in terms of the initially given parameters.
B) Post-collision phase:
Only gravity does work on the system here, due to the no-slip condition.
Hence, mechanical energy is conserved.

Good luck&good constitution day to all Spaniards!..:smile:
 
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}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top