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A Sphere rolling down an incline. Find the speed at the bottom

  • Thread starter kavipach
  • Start date
  • #1
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Homework Statement


A sphere of radius r_0 = 24.0 cm and mass m = 1.20{\rm{ kg}} starts from rest and rolls without slipping down a 37.0^\circ incline that is 15.0 m long.
Calculate its translational speed when it reaches the bottom.

Homework Equations


ke=1/2mv^2
pe=mgh

The Attempt at a Solution


Since PE=KE , i put the two equations together to form mgh=1/2mv^2. Canceling the m's we get gh=1/2v^2. Using triganomity i find that the height is 9.027m. Then pluging in all the values the answer comes out to be 13.3m/s. This isnt the right answer.
 

Answers and Replies

  • #2
alphysicist
Homework Helper
2,238
1
Hi kavipach,

Homework Statement


A sphere of radius r_0 = 24.0 cm and mass m = 1.20{\rm{ kg}} starts from rest and rolls without slipping down a 37.0^\circ incline that is 15.0 m long.
Calculate its translational speed when it reaches the bottom.

Homework Equations


ke=1/2mv^2
pe=mgh

The Attempt at a Solution


Since PE=KE , i put the two equations together to form mgh=1/2mv^2.
Your equation is neglecting the kinetic energy due to the rolling motion of the sphere. Once you add the rotational kinetic energy term you should get the right answer.
 
  • #3
As mentioned above, what you have solved for is the situation of a sliding mass, not a rolling mass. In the sliding case, all of the kinetic energy of the mass is in its translational motion, so [tex]K_E[/tex] is simply [tex]K_E = K_{translational} = (1/2)mv^2[/tex], which is the equation you used.

However, in the rolling case, the kinetic energy of the mass is in its translational *and* rolling motion. Then your equation will become:

[tex]P_E = K_E [/tex]
[tex]P_E = K_{translational} + K_{rotational}[/tex]

The potential energy is now shared between the translational and rotational kinetic energies, so the translational speed at the bottom will be less than that of a sliding mass.

To actually solve this, you will need to know the equation for rotational kinetic energy and for the moment of inertia of a solid sphere... sound familiar?
 

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