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10-22-2005
03:00 AM

10-22-2005
03:00 AM

rotational kinetic energy

hey, Im hoping somebody can lend me some help. I'm having extreme difficulties on what should be a pretty straight forward question. Here it is:

A bowling ball of mass 7.3kg. and radius 9.0m rolls without slipping down a lane at 3.3m/s. claculate its total kinetic energy.

So first thing I did was to obtain w= v/r

w= 36.67 rad/s

then I believe the formula/eq'n that is sapsoe to be used should be

EK= 1/2 mv^2 + 1/2 Iw^2

,where I=MR^2

another thing I'm not sure to account for is the momnet of inertia of the sphere which would be I= (2/5) MR^2

so, I plugged in my values and I get a large value, about 80J if I dont account for the (2/5) of I and if I don't then I get about 32J. Both obviously incorrect.

The answer given is, 56J

Okay thanks a lot for the help!

A bowling ball of mass 7.3kg. and radius 9.0m rolls without slipping down a lane at 3.3m/s. claculate its total kinetic energy.

So first thing I did was to obtain w= v/r

w= 36.67 rad/s

then I believe the formula/eq'n that is sapsoe to be used should be

EK= 1/2 mv^2 + 1/2 Iw^2

,where I=MR^2

another thing I'm not sure to account for is the momnet of inertia of the sphere which would be I= (2/5) MR^2

so, I plugged in my values and I get a large value, about 80J if I dont account for the (2/5) of I and if I don't then I get about 32J. Both obviously incorrect.

The answer given is, 56J

Okay thanks a lot for the help!

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2 REPLIES 2

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10-22-2005
03:00 AM

10-22-2005
03:00 AM

rotational kinetic energy

On 10/22/2005 12:52:38 PM, mdao wrote:

>hey, Im hoping somebody can

>lend me some help. I'm having

>extreme difficulties on what

>should be a pretty straight

>forward question. Here it is:

>

>A bowling ball of mass 7.3kg.

>and radius 9.0m rolls without

>slipping down a lane at

>3.3m/s. claculate its total

>kinetic energy.

>

>So first thing I did was to

>obtain w= v/r

>w= 36.67 rad/s

>

>then I believe the

>formula/eq'n that is sapsoe to

>be used should be

>

>EK= 1/2 mv^2 + 1/2 Iw^2

>

>,where I=MR^2

>

>another thing I'm not sure to

>account for is the momnet of

>inertia of the sphere which

>would be I= (2/5) MR^2

>

>so, I plugged in my values and

>I get a large value, about 80J

>if I dont account for the

>(2/5) of I and if I don't then

>I get about 32J. Both

>obviously incorrect.

>

>The answer given is, 56J

>Okay thanks a lot for the

>help!

Perhaps rechecking your math is in order. Using the same inputs and equations, I got 55.6J.

Your rotational speed seems WAY TOO high (3.3m/s)/9m is less than 1 rad/s. The circumference is 56.5m. You cannot possibly get more than 1 rotation/s. Your rotational energy calculation seems to in question. With that large an angular speed, you cannot get only 48 J.

Your linear energy is also suspect.1/2mv^2 comes out to 39.7J on my calculator.

TTFN,

Eden

>hey, Im hoping somebody can

>lend me some help. I'm having

>extreme difficulties on what

>should be a pretty straight

>forward question. Here it is:

>

>A bowling ball of mass 7.3kg.

>and radius 9.0m rolls without

>slipping down a lane at

>3.3m/s. claculate its total

>kinetic energy.

>

>So first thing I did was to

>obtain w= v/r

>w= 36.67 rad/s

>

>then I believe the

>formula/eq'n that is sapsoe to

>be used should be

>

>EK= 1/2 mv^2 + 1/2 Iw^2

>

>,where I=MR^2

>

>another thing I'm not sure to

>account for is the momnet of

>inertia of the sphere which

>would be I= (2/5) MR^2

>

>so, I plugged in my values and

>I get a large value, about 80J

>if I dont account for the

>(2/5) of I and if I don't then

>I get about 32J. Both

>obviously incorrect.

>

>The answer given is, 56J

>Okay thanks a lot for the

>help!

Perhaps rechecking your math is in order. Using the same inputs and equations, I got 55.6J.

Your rotational speed seems WAY TOO high (3.3m/s)/9m is less than 1 rad/s. The circumference is 56.5m. You cannot possibly get more than 1 rotation/s. Your rotational energy calculation seems to in question. With that large an angular speed, you cannot get only 48 J.

Your linear energy is also suspect.1/2mv^2 comes out to 39.7J on my calculator.

TTFN,

Eden

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10-22-2005
03:00 AM

10-22-2005
03:00 AM

rotational kinetic energy

A bowling ball with a nine meter radius? Exactly who's bowling? Zeus?

Did you perhaps mean nine centimeters? That would be more consistent with your angular velocity. However it turns out that your radius does not matter -- the answer is independent of the radius, and there is no reason why the problem should actually state a radius.

I=m·r² applies only where the entire mass is at a distance r from the rotation axis. This could be a point mass, or a uniform ring, or a cylinder. But it does not apply to a ball, where only the the equator is at a distance r from the rotational axis. The moment of inertia of a ball is as you give it,(2/5)·m·r² . And that is what you must use.

Your value of about 80J would arise if you used the incorrect value for the moment of inertia, the one for a ring or cylinder. Your value of about 32J would arise if you attempted to apply the factor of 2/5 to the total kinetic energy. That factor applies only to the moment of inertia, hence the rotational kinetic energy.

� � � � Tom Gutman

Did you perhaps mean nine centimeters? That would be more consistent with your angular velocity. However it turns out that your radius does not matter -- the answer is independent of the radius, and there is no reason why the problem should actually state a radius.

I=m·r² applies only where the entire mass is at a distance r from the rotation axis. This could be a point mass, or a uniform ring, or a cylinder. But it does not apply to a ball, where only the the equator is at a distance r from the rotational axis. The moment of inertia of a ball is as you give it,

Your value of about 80J would arise if you used the incorrect value for the moment of inertia, the one for a ring or cylinder. Your value of about 32J would arise if you attempted to apply the factor of 2/5 to the total kinetic energy. That factor applies only to the moment of inertia, hence the rotational kinetic energy.

� � � � Tom Gutman