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PHYSICAL SCIENCE IS IMAGINATIVE

SO THINK

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2006-08-20 02:44:29 · 10 answers · asked by Anonymous in Science & Mathematics Physics

10 answers

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I THINK IT WILL BE

1 / 3of its kinetic energy is rotational

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2006-08-20 03:02:14 · answer #1 · answered by Anonymous · 3 1

1) If Kt is translational kinetic energy, it equals:

Kt=1/2 mv^2
where m is mass and v is velocity

2) If Kr is rotational kinetic energy, it equals:

Kr=1/2 Iw^2
where I is moment of inertia and w is angular velocity

3) I = mr^2 where m is mass and r is the radius of the cylinder

w=v/r where v is velocity of the surface of the cylinder and r is the radius.

If the cylinder is not slipping, then the point touching the ground is stationary at the instant it's touching the ground. That means the speed of the surface of the cylinder equals the translational velocity of the cylinder; it's just going the opposite direction.

In other words, the translational velocity of the cylinder is all you need.

The total kinetic energy equals the translational and the rotational kinetic energy. Substituting, using the equations from 3), you get the total kinetic energy in terms of mass, radius, and velocity:

K = 1/2 mv^2 + 1/2 (mr^2) (v/r)^2

K = 1/2 mv^2 + 1/2 (mr^2) (v^2/r^2)
The radius cancels out, since you have r^2 in both numerator and denominator:

K = 1/2 mv^2 + 1/2 mv^2
The translational kinetic energy is the left term and the rotational kinetic energy is the right term.

To find the percentage of kinetic energy that is rotational, divide the rotational kinetic energy by the total kinetic energy:

(1/2 mv^2) / (1/2 mv^2 + 1/2 mv^2)
Obviously, half of the kinetic energy is due to rotational motion.

2006-08-20 03:12:31 · answer #2 · answered by Bob G 6 · 2 0

As the cylinder it has linear and rotational kinetic energies.

E(k-total)=E(k-linear) + E( k-rotational)

E(k-total)=(mv^2)/2 +(Iw^2)/2

I=(mh^2)/12 +(mr^2)4
m-mass of the cylinder
r- radius
h- height

v=2Pi r/t linear speed
w=v/r rotational speed

We have
E(k-total) =
= (mv^2)/2 + ((mh^2)/12 +(mr^2)4)(v/r)^2 =
= mv^2 (1+ (h^2)/12 + 1/4)=
=mv^2(15+ h^2)/12)

2006-08-20 04:45:45 · answer #3 · answered by Edward 7 · 1 0

i am going to flow right into a lot more beneficial ingredient in case you opt for, even if the quick answer isn't any. both sturdy uniform cylinders might want to attain the bottom of the prone airplane mutually, assuming they roll without slipping.

2016-11-05 05:33:58 · answer #4 · answered by Anonymous · 0 0

100%
no slipping means NO Friction, therefore, 100% of the energy is being used for movement.

2006-08-20 02:50:12 · answer #5 · answered by FLSTC 2 · 0 1

Two types of motion involved here:
the forward motion and rotational now u can work it out.

http://voip.helpsiteonline.com/

2006-08-20 03:00:52 · answer #6 · answered by Bulldog 3 · 0 0

It all depends upon the raius of the cylinder, otherwise it is impossible.

2006-08-20 03:40:41 · answer #7 · answered by Cool Solve 2 · 0 0

wow.

it should be its potential energy - Kinetic energy divided by Pi correct?



i actually can't remember how to do this question...ugh..it's been so long.

2006-08-20 02:50:53 · answer #8 · answered by smartyphred 2 · 0 1

100%, as it is rolling without slipping :-)

2006-08-20 02:51:02 · answer #9 · answered by DG 3 · 0 1

if it is level it will decrease as it deccelerates. but it is 50% and gravity is other 50%

2006-08-20 02:55:08 · answer #10 · answered by Boliver Bumgut 4 · 0 0

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