Hollow sphere, angular momentum, torque problem? help?

[nchin]

Hollow sphere, angular momentum, torque problem? help??

a hollow sphere or radius 0.15m with rotational inertia = 0.040 kg m^2 about a line through its center of mass, rolls without slipping up a surface inclined 30 degree to the horizontal. at a certain initial position,the sphere's total kinetic energy is 20J.

a) How much of this initial KE is rotational?

what i did

Hollow sphere = 2/3MR^(2)
KErot = 1/2 Iω^2

KEtotal = KEtranslational + KErotational

a) 20 = 1/2mv^2 + 1/2 x 2/3 mr^2
20 = 1/2mv^2 + 1/3 mr^2

this part confuses me
1/2mv^2 + 1/3 mr^2 = 5/6mv^2
20 = 5/6mv^2

How was 1/2mv^2 + 1/3 MR^2 = 5/6mv^2?? mv and MR are different yet how were they both added together??

 

[haruspex]

When dimensions don't match up, backtrack to the first instance. It's here:

a) 20 = 1/2mv^2 + 1/2 x 2/3 mr^2

ω is missing.

 

[nchin]

When dimensions don't match up, backtrack to the first instance. It's here:

ω is missing.

1/2mv^2 + 1/2 x 2/3 mr^2ω = 5/6mv^2

ok so if i put a ω there, how would it still = 5/6mv^2? in other words how can you change mr^2ω into mv^2? help!

 

[nchin]

oh nvm i got it. w = V^2/R^2. r cancel out. thanks!

 

[Nickie]

The reason for this is because in this problem, we are dealing with a rolling motion of the hollow sphere. This means that both linear and rotational motion are present and they are related to each other. In order to find the total kinetic energy, we need to take into account both types of motion.

The translational kinetic energy is given by 1/2mv^2, where m is the mass of the sphere and v is its linear velocity. The rotational kinetic energy is given by 1/2Iω^2, where I is the moment of inertia and ω is the angular velocity.

Now, for a rolling motion, the linear velocity and angular velocity are related by v = ωR, where R is the radius of the sphere. Therefore, we can substitute this into the equations for translational and rotational kinetic energy to get:

KEtranslational = 1/2mv^2 = 1/2m(ωR)^2 = 1/2mω^2R^2
KErotational = 1/2Iω^2 = 1/2(2/3mr^2)ω^2 = 1/3mr^2ω^2

Adding these two equations together, we get:

KEtotal = KEtranslational + KErotational = 1/2mω^2R^2 + 1/3mr^2ω^2

Now, we can factor out ω^2 from both terms and we get:

KEtotal = (1/2m + 1/3mr^2)ω^2

Since we know that the total kinetic energy is 20J, we can set this equal to the above equation and solve for ω^2:

20 = (1/2m + 1/3mr^2)ω^2
ω^2 = (20/(1/2m + 1/3mr^2))

Now, we can substitute this value for ω^2 back into the equations for translational and rotational kinetic energy to get:

KEtranslational = 1/2m(ωR)^2 = 1/2m(20/(1/2m + 1/3mr^2))R^2 = 10R^2/(1/2m + 1/3mr^2)
KErotational = 1/2Iω^