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April 18, 2014

April 18, 2014

Posted by **Kayla** on Sunday, July 15, 2012 at 8:46pm.

- Physics -
**bobpursley**, Sunday, July 15, 2012 at 10:51pmWhat is the translational (1/2 mv^2) PLUS the initial rotational energy (1/2 I w^2, where w=vr). Will that equal mgh?

- Physics -
**Kayla**, Sunday, July 15, 2012 at 11:08pmThanks bobpursley! I was somewhere close to that idea. I had mgh=1/2mv^2 + 1/5mv^2. I let h=12sin30. I was solving for velocity for some reason, doh!

- Physics -
**Kayla**, Sunday, July 15, 2012 at 11:17pmso I have mgh = 300.

i have my rotational energy = 350.

so it will reach the top, correct?

- Physics -
**Elena**, Monday, July 16, 2012 at 1:06amTry to obtain the general solution of the problem, i.e.,

KE=PE

KE(tr) + KE(rot) =PE

0.7m•v²/2 = m•g•h

h=0.7•v²/g=70/9.8=7.14 m.

h(real)=s•sinα=12•0.5=6 m

Since 7,14>6, the ball‘ll

rich the top of the ramp.

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