Calculating Velocity at the Bottom of a Quarter Circle Track

Click For Summary
SUMMARY

The discussion centers on calculating the velocity of a block with a mass of 10 kg at the bottom of a quarter circle track with a radius of 10 m, given an initial downward velocity of 10 m/s. The relevant equations include the conservation of energy, specifically the kinetic and potential energy formulas. The correct final velocity at the bottom of the track is determined to be 17.2 m/s. The participants clarify the relationship between the radius of the track and the height of the ramp, confirming that they are equivalent in this scenario.

PREREQUISITES
  • Understanding of kinetic energy (KE) and potential energy (PE) concepts
  • Familiarity with the conservation of mechanical energy principle
  • Knowledge of rotational dynamics, including moment of inertia (I)
  • Basic understanding of circular motion and radius-height relationships
NEXT STEPS
  • Study the conservation of energy in mechanical systems
  • Learn about moment of inertia and its applications in rotational motion
  • Explore the relationship between radius and height in circular tracks
  • Investigate the effects of friction on energy conservation in real-world scenarios
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and energy conservation, as well as educators seeking to clarify concepts related to rotational and translational motion.

tennisgirl92
Messages
43
Reaction score
1

Homework Statement


A block of mass 10kg starts at the top of a frictionless track which forms a quarter circle with radius 10m. It is given an initial downward velocity of 10m/s. What is the velocity at the bottom of the track?

Homework Equations


KEtranslational, f+ KErotational, f+PEf=KEtranslational, i+KErotational, i+ PEf
KE, rotational=1/2Iw2
KE, translational=1/2mv2
PE=mgh
I=MR2
w=v/r

The Attempt at a Solution


I believe the initial equation would be set up like this.
1/2mvf2+ 1/2 MR2(V/R)2+mghf=1/2mvi2+1/2 MR2(V/R)2+mghi

which would then reduce to:
vf2=vi2+ghinitial

This does not seem like the likely way to do it because we are not given the height of the mass at the top. Should I be using the rotational and translational KE in the equation? At what point is it rotational and what point translational?
 
Physics news on Phys.org
tennisgirl92 said:
vf2=vi2+ghinitial

This does not seem like the likely way to do it because we are not given the height of the mass at the top.

Look carefully.

tennisgirl92 said:
Should I be using the rotational and translational KE in the equation? At what point is it rotational and what point translational?

Why would it rotate? When does something rotate?
 
Mastermind01 said:
Look carefully.Why would it rotate? When does something rotate?

Would the radius be the initial height?

When it is rolling? I'm not exactly sure.
 
tennisgirl92 said:
Would the radius be the initial height?

That is correct.

tennisgirl92 said:
When it is rolling? I'm not exactly sure.

Try and think intuitively if you don't know the physics behind it.
 
Mastermind01 said:
That is correct.
Try and think intuitively if you don't know the physics behind it.
Ok, now I see how to do this. I obtained 17.2 m/s, which is correct. Thank you!

However, I still don't see how the radius could be equal to the height of the ramp. There is a picture provided that looks like a skateboard ramp, with a strict vertical height and then the inside being curved with a radius. The vertical height does not look equal to the radius, even though I know this is not to scale.
 
Does the picture below look like the picture you have of the skateboard ramp? Is the quarter circle oriented like that?
Quarter circle track.jpg
 
  • Like
Likes   Reactions: Mastermind01
TomHart said:
Does the picture below look like the picture you have of the skateboard ramp? Is the quarter circle oriented like that?View attachment 197493
yes, it is just like that, except flipped so that the curve is on the right and height on the left.
 
So does it make sense to you that the radius = height?
 
Yes, I do see it now-my picture did not have the radius being pointed horizontally and vertically, just down the middle. Thank you for all your help!
 
  • Like
Likes   Reactions: TomHart

Similar threads

  • · Replies 12 ·
Replies
12
Views
12K
  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 11 ·
Replies
11
Views
6K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
21
Views
3K
Replies
9
Views
3K
Replies
4
Views
1K
  • · Replies 17 ·
Replies
17
Views
4K