Speed of Objects Rolling and Sliding Down a Hill

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Homework Help Overview

The problem involves a solid uniform marble and a block of ice, both starting from rest at the same height on a hill. The marble rolls without slipping while the ice slides without friction. Participants are discussing the speeds of both objects when they reach the bottom of the hill.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the application of energy conservation principles to find the speeds of the marble and ice. There are attempts to derive equations for both objects, with some questioning the correctness of their calculations and the interpretation of variables.

Discussion Status

Some participants have confirmed the correctness of the approach for the block of ice, while others are working through the derivation for the marble. There is acknowledgment of errors in calculations, and one participant has identified a potential issue with variable notation that affected their solution.

Contextual Notes

Participants are navigating issues related to variable representation and the implications of mass in their equations. There is a focus on ensuring the correct application of energy conservation without friction for the ice and considering rotational inertia for the marble.

valeriex0x
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Homework Statement



A solid uniform marble and a block of ice, each with the same mass, start from rest at the same height H above the bottom of a hill and move down it. The marble rolls without slipping, but the ice slides without friction.

Find the speed of each of these objects when it reaches the bottom of the hill.
Vice:?
Vmarble:?

Homework Equations



1/m v^2 = mgh

The Attempt at a Solution



Ice:
1/2 mv^2=mgh

v= squareroot 2gh

Marble:
mgh=1/2 mv^2 + 1/2 I ω^2

=1/2 mv^2 +1/2(2/5 mr^2) (v/r)^2

The answers i entered in were marked incorrect by masteringphysics.

Please advise! : (
 
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Hi valerie! x0x :smile:

Your v for the block of ice is correct.
And your equation for the marble is correct too.

The only thing missing is that you didn't give v of the marble.
Did you solve your equation for v?
 
well i got stuck... and i don't know why its marking me wrong. for the marble i got v: squareroot 2/4 mgh


anythoughts?
 
valeriex0x said:
well i got stuck... and i don't know why its marking me wrong. for the marble i got v: squareroot 2/4 mghanythoughts?

I'm afraid that in that case you did not solve your own equation correctly.

For starters, there should not be an "m" in there.
There isn't an "m" for the block of ice either!

But your fraction of 2/4 is wrong too.

How did you arrive at your solution for v?
Can you show some (major) steps?
 
oh never mind ... this stupid mastering physics wanted a capital H instead of h.

aughhhhhh! the marble was squareroot 10 gh/7
 
thanks for your help! i caught that at the last min!
 
Congrats with your right answer! :smile:
 

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