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Kinetic and Potential energy problem

  • Thread starter BrainMan
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  • #26
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I get 2.08 m/s. The ball is 2.2 cm in radius, yes?
But I agree, the book answer is way off. I can't see what simple error leads to it.
I don't understand your reasoning. Even at the lowest point, the ball is rotating as well as moving horizontally. However, it turns out that the energy taken up by that is only 0.01% of the total, so it can be safely ignored. It would be a very different story if the radius were similar in magnitude to the length of the string.
woops you're right, hit square root instead of cube root I think. As I said, I'm no expert. When I did these kinds of problems in my mechanics class we never took rotation into account. If we took rotation into account, wouldn't the energy be the exact same anyways?
## \frac {1}{2} Iω^2 = \frac {1}{2}(mr^2) \frac {v^2}{r^2} = \frac {1}{2} mv^2## no?
 
  • #27
haruspex
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If we took rotation into account, wouldn't the energy be the exact same anyways?
## \frac {1}{2} Iω^2 = \frac {1}{2}(mr^2) \frac {v^2}{r^2} = \frac {1}{2} mv^2## no?
No, you're forgetting the length of the string. ##\omega = \frac v{l+r}##. The total KE is ## \frac {1}{2} Iω^2 + \frac {1}{2}mv^2 = \frac {1}{2} \frac 25 m r^2 ω^2 + \frac {1}{2}mv^2 = \frac {m}{2} v^2(\frac 25 \left(\frac r{l+r}\right)^2 + 1) ##
 
  • #28
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No, you're forgetting the length of the string. ##\omega = \frac v{l+r}##. The total KE is ## \frac {1}{2} Iω^2 + \frac {1}{2}mv^2 = \frac {1}{2} \frac 25 m r^2 ω^2 + \frac {1}{2}mv^2 = \frac {m}{2} v^2(\frac 25 \left(\frac r{l+r}\right)^2 + 1) ##
Lolwut. I've done like maybe 30 problems involving torque or angular velocity if we discount my waves course. Is there a good link that you know of about this relationship? I have too many questions to reasonably ask you to answer like: Isn't the radius the distance from the center of mass to the point that we're rotating about so r in this case accounts for the length of the string? Where did the 2/5 come from? What's the ##(\frac {r}{l+r})^2##? Where did the extra v^2 come from on the angular energy in the last equation. Aren't we counting the kinetic energy twice?

Don't bother trying to answer all of that, I'll look it all up after finals are over.
 
  • #29
haruspex
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Lolwut. I've done like maybe 30 problems involving torque or angular velocity if we discount my waves course. Is there a good link that you know of about this relationship? I have too many questions to reasonably ask you to answer like: Isn't the radius the distance from the center of mass to the point that we're rotating about so r in this case accounts for the length of the string? Where did the 2/5 come from? What's the ##(\frac {r}{l+r})^2##? Where did the extra v^2 come from on the angular energy in the last equation. Aren't we counting the kinetic energy twice?

Don't bother trying to answer all of that, I'll look it all up after finals are over.
You can treat the motion of the sphere as either a rotation about the support point of the string or as a linear motion of the sphere's centre plus a rotation about that centre.
The angular speed is the same either way. The linear speed of the sphere's centre is v = (l+r)ω.
The moment of inertia of a uniform solid sphere mass M radius r about its centre is 2Mr2/5. Standard formula.
If we treat it as a linear motion plus a rotation then we must add the KEs:
##\frac 12 \frac 25 M r^2 \omega^2 + \frac 12 M v^2 = \frac 12 M (\frac 25 \frac {r^2 }{(l+r)^2} v^2 + v^2) = \frac 12 M v^2(\frac 25 \frac {r^2 }{(l+r)^2} +1) ##
Treating it as a rotation about the support point, we need the moment of inertia about that point. For this we use the parallel axis theorem, which tells us we have to add the mass of the sphere times the square of the displacement, giving 2Mr2/5 + M(l+r)2.
##\frac 12 (\frac 25 M r^2 + M(l+r)^2)\omega^2 ## gives the same result.
 
  • #30
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You can treat the motion of the sphere as either a rotation about the support point of the string or as a linear motion of the sphere's centre plus a rotation about that centre.
The angular speed is the same either way. The linear speed of the sphere's centre is v = (l+r)ω.
The moment of inertia of a uniform solid sphere mass M radius r about its centre is 2Mr2/5. Standard formula.
If we treat it as a linear motion plus a rotation then we must add the KEs:
##\frac 12 \frac 25 M r^2 \omega^2 + \frac 12 M v^2 = \frac 12 M (\frac 25 \frac {r^2 }{(l+r)^2} v^2 + v^2) = \frac 12 M v^2(\frac 25 \frac {r^2 }{(l+r)^2} +1) ##
Treating it as a rotation about the support point, we need the moment of inertia about that point. For this we use the parallel axis theorem, which tells us we have to add the mass of the sphere times the square of the displacement, giving 2Mr2/5 + M(l+r)2.
##\frac 12 (\frac 25 M r^2 + M(l+r)^2)\omega^2 ## gives the same result.
Oooh ok. Ya that makes sense, tried to understand moment of innertia off a quick wiki read. mr2 isn't the moment of innertia. I just read the link you posted up top about innertia. I assumed it was the radius of the circle it's rotating around. I haven't ever even heard of the parallel axis theorem, I'll give it a look after finals but the other part made perfect sense. Thank you.
 

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