Solving a 2m Inclined Plane Problem with Negligible Friction

[cortozld]

Homework Statement

A solid ball is at the top of a 2 meter inclined plane. Assuming friction is negligible, what is its linear speed (m/s) as it reaches the bottom?

Homework Equations

PE=mgh
KE=.5mv^2
RKE=.5Iw^2
I of ball=.4mr^2
w=v/r

The Attempt at a Solution

Pretty much what i did was set PE=KE+RKE
which is mgh=.5mv^2+.5(.4mr^2)+w^2 masses cancel out and so does r when v/r is replaced with w

equation is now gh=.5v^2+.5(.4)v^2
if i solve for v i get: 2v^2=2gh/.4= v^2=gh/.4

thus v=sqrt(9.81*2/.4)

v=7.0 m/s

i really have no idea if I am doing this right...

 

[kuruman]

In such problems, I find statements like "assume that friction is negligible" confusing.

If friction is really negligible, then the ball slides down the incline without rolling. In that case, it might as well be replaced by a sliding block and there is no ω.

If by "assume that friction is negligible" the intent is to indicate that the ball is actually rolling without slipping but frictional losses are negligible (i.e. mechanical energy is conserved), then the problem should state so more clearly.

Your method is correct if you assume that the latter interpretation applies.

 

[cortozld]

Yes, your second statement was correct. As for the problem not being specific it's my teacher not me so talk to him :). As for the answer apparently mine is wrong. So if you have any ideas about something I've missed I'd be glad to hear it

 

[kuruman]

equation is now gh=.5v^2+.5(.4)v^2
if i solve for v i get: 2v^2=2gh/.4= v^2=gh/.4

The first line is correct. Can you show me how you get from the first line to v²=gh/0.4?

 

[cortozld]

From gh=.5v^2+.5(.4)v^2 I multiplied both sides by 2 to get rid of the .5: 2gh=v^2+.4v^2. Then I divided by .4 and added the v^2s together: 2gh/.4=2v^2. Divided both sides by 2 again to get: gh/.4=v^2.

 

[kuruman]

From gh=.5v^2+.5(.4)v^2 I multiplied both sides by 2 to get rid of the .5: 2gh=v^2+.4v^2. Then I divided by .4 and added the v^2s together: 2gh/.4=2v^2. Divided both sides by 2 again to get: gh/.4=v^2.

Not correct. First you add the v², then you divide by whatever multiplies the v². In other words, you factor out the v² term to get v²(1+0.4).

 

[cortozld]

So that means i have 2gh=v^2(1+.4). Then divide by (1+.4): 2gh/(1+.4)=v^2 and vinally solve for v. which is v=sqrt(2*9.81*2)/(1+.4)

so v=5.3 m/s! Thanks a lot :)