Rotational motion- potential and kinetic energy

AI Thread Summary
The discussion focuses on finding the linear velocity of a sphere as it leaves a cliff using conservation of energy principles. The moment of inertia for a sphere is noted, and the user questions the need to account for friction in the context of rolling without slipping. It is clarified that static friction does not perform work since there is no relative motion, meaning energy is conserved in this scenario. The user correctly sets the initial kinetic and potential energies equal to the final energies, concluding that friction does not need to be included in the calculations. The overall consensus is that the approach to solving the problem is valid without factoring in friction.
oxymoron man
Messages
1
Reaction score
0

http://i.imgur.com/0RtN9Ui.png?1

I am trying to find the linear velocity of the sphere as it leaves the cliff.

2. Homework Equations :
Moment of inertia of a sphere: I= 2/5 mr2

3. My attempt at a solution:

I know that I am supposed to solve this through the conservation of energy, but would I not have to account for friction somehow since it is rolling without slipping? I know how to find the kinetic and potential energies at both points but the solutions manual (i know, i know, i cheated) says nothing about accounting for the non-conservative force of friction that is necessary for rolling without slipping.
Either way, without friction I have

U-initial: 0
K-initial= 1/2 mv^2+(1/2)*(2/5*m*r^2)*v^2/r^2
k-initial= 1/2 mv^2+1/5 mv^2
k-initial: 7/10 mv^2

and for
U-final: mgh
K-final: 7/10 mvfinal^2

Could i just set these equal to each other and get the answer? No friction needed?
 
Physics news on Phys.org
oxymoron man said:
http://i.imgur.com/0RtN9Ui.png?1

I am trying to find the linear velocity of the sphere as it leaves the cliff.

2. Homework Equations :
Moment of inertia of a sphere: I= 2/5 mr2

3. My attempt at a solution:

I know that I am supposed to solve this through the conservation of energy, but would I not have to account for friction somehow since it is rolling without slipping? I know how to find the kinetic and potential energies at both points but the solutions manual (i know, i know, i cheated) says nothing about accounting for the non-conservative force of friction that is necessary for rolling without slipping.
Either way, without friction I have

U-initial: 0
K-initial= 1/2 mv^2+(1/2)*(2/5*m*r^2)*v^2/r^2
k-initial= 1/2 mv^2+1/5 mv^2
k-initial: 7/10 mv^2

and for
U-final: mgh
K-final: 7/10 mvfinal^2

Could i just set these equal to each other and get the answer? No friction needed?
You're doing fine. Since the sphere is rolling without slipping, only static friction is involved. Static friction does no work because there's no relative motion. You start with some initial KE and lose some to gravitational PE as you've stated. No energy is lost due to friction.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top