A solid sphere of uniform density starts from rest and rolls without slipping a

Question

A solid sphere of uniform density starts from rest and rolls without slipping a distance of d = 3.3 m down a q = 28° incline. The sphere has a mass M = 3.1 kg and a radius R = 0.28 m.

My question is:

What is the translational kinetic energy of the sphere when it reaches the bottom of the incline?

KEtran =

The equation i used was: 1/.7145 (translational fraction) * 9.8*distance*sin(28)* .5

I keep getting the wrong answer, if you can provide what I am doing wrong with an answer as well I would much appreciate it, thanks!

Explanation / Answer

In order to find the translational kinetic energy you have to find the total kinetic energy of the object so your first step should be to setup the initial and final energies of the object. First let's look at the top where it starts (initial energy or Eo) It starts from rest so that means it will only have gravitational potential energy to begin or (m)(g)(h). We know the mass and acceleration due to gravity so lets find the height of the object. =28o and since we are attempting to find the side opposite to it lets use sin()= opposite/hypotenuse. The hypotenuse is 3.3m so the opposite side must be 3.3sin(28). So Eo=(3.1)(9.8)(3.3)(sin(28))47.0664. Now we have to find the final energy, but keep in mind it will have translational and rotational kinetic energy at the bottom so Ef=(1/2)(m)(vt2)+(1/2)(I)(2). Now let's manipulate the 2nd half of the equation for Ef using the identity =(v/r) and the equation for the moment of Inertia for a solid sphere I=(2/5)(m)(r2). Using these identities Ef=(1/2)(m)(vt2)+(1/2)(2/5)(m)(r2)(vt2/r2)

Now simplify the equation: (1/2)(m)(vt2)+(1/5)(m)(vt2) or (7/10)(m)(vt2) Now set Eo=Ef to solve for vt.

mgh=(7/10)(m)(vt2) :simplify the masses cancel out from both sides, multiply both sides by (10/7) and then take the square root of both sides: [(10/7)gh]=vt => [(10/7)(9.8)(3.3)(sin(28))]4.65721(m/s). Now use vt to solve for the translational kinetic energy of the solid sphere. (1/2)(3.1)(4.657212)33.6189J using 2 significiant figures Ekt=34J and using 3 significant figures Ekt=33.6J.

Here's what you did wrong you found the height and then multiplied it times gravity , but then you mutiplied that times some number. You have to use the law of conservation of energy for this problem (Eo+|Wnc|=Ef) and the sign of |Wnc| is determined by whether the work is adding energy(+) (i.e a force in the direction of motion) or subtracting energy(-) (i.e a force like friction that works against the direction of motion).

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