A Rolling Disk

  • Thread starter kenricktan
  • Start date
  • #1

Homework Statement


A solid, Uniform disk of radius 5.00 cm and mass 1.50 kg is rolling without slipping a long a horizontal surface. The disk makes 2.00 revolution per second

a. Find the total kinetic energy (translational + rotational) of the disk

b. Find the minimum height h of the step (placed in front of the rolling disk) that will prevent the disk from rolling past it. (Hint: assume that the hight h is adjusted so that the disk rolls just up to the top of the step and stops. Conserve Energy)


Homework Equations


W= 2.00 Revolution x 2pi radian
V = wr
I = (1/2)mr^2
KE_rot = (1/2)Iw^2
KE = (1/2)mv^2


The Attempt at a Solution


a. I assumed that K_total = KE_rot + KE
K_total = 4.46 x 10^-3 Joules

b. I have no Idea how to solve this....help?
 

Answers and Replies

  • #3
67
4
The height must be such that energy will be conserved between the initial translational and rotational velocity and the final trans/rot velocity. Ktotal(0) + U(0) = Ktotal(1) + U(1)

I'd begin by drawing a picture. What does the final total Kinetic energy have to be?
 
  • #4
I tried using 1/2 m Vo^2 + mgh(o) = 1/2 m V1^2 + mgh(1)

and since it says that assuming that it "stops" at the top of the step so I set V1 = 0
and also h(o) = 0

cancels all of the m
so I get h1 = Vo^2 / 2g
Is that right?
 
  • #5
67
4
I tried using 1/2 m Vo^2 + mgh(o) = 1/2 m V1^2 + mgh(1)

OK let's stop right here for a sec. 1/2 * Vo^2 is the translational kinetic energy, but you also need to include the rotational kinetic energy because the object is rotating as well as moving forward.
 

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