Prob. 51 Physics: Moment of Inertia for 10 kg Rolling Cylinder

In summary, the problem involves a 10 kg cylinder rolling without slipping on a horizontal surface with a speed of 10 m/s at its centre of mass. The translational kinetic energy of its centre of mass is 500 J, and the total energy is the sum of its translational and rotational kinetic energy. However, the rotational kinetic energy is unknown and is equal to 250 J. The condition of "rolling without slipping" gives the relationship between translational and rotational speeds. The moment of inertia for a cylinder is needed to calculate the rotational kinetic energy.
  • #1
salehajaweid
4
0
Hi I have a few problems from 'Physics for Scientists and Engineers' by Serwat and Jewett, chap 10

Prob. 51

A cylinder of mass 10 kg rolls without slipping on a horizontal surface. At the instant its centre of mass has a speed of 10 m/s. Determine (a) the translational kinetic energy of its centre of mass (b) the rotational kinetic energy about its centre of mass (c) its total energy.

(a) turns out to be 500 J and (c) is the sum of E.K. rotational and E.K. translational however (b) is a mystery for me, I would appreciate someone helping me out.

Thanks.
 
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  • #2
Hint: What does "rolling without slipping" mean?
 
  • #3
Is that a hint or a question? If it's a hint I don't get the connection ... if it's a ques well in the rolling without slipping scenario velocity of the contact point (with the surface) is zero ... I know that if v of centre of mass will be half of the v at the top point. However for rotational kinetic energy I would need the moment of inertia, how do i calculate that? Btw the answer is 250 J, in case tht helps
 
  • #4
salehajaweid said:
Is that a hint or a question?
Sure it's a hint! :smile:
If it's a hint I don't get the connection ... if it's a ques well in the rolling without slipping scenario velocity of the contact point (with the surface) is zero ... I know that if v of centre of mass will be half of the v at the top point.
The condition for "rolling without slipping" gives you the relationship between translational and rotational speeds.
However for rotational kinetic energy I would need the moment of inertia, how do i calculate that?
What's the moment of inertia of a cylinder? :wink:
 
  • #5
Thanks Doc Al! :)
 

1. What is the formula for calculating the moment of inertia for a rolling cylinder?

The formula for calculating moment of inertia for a rolling cylinder is I = (1/2)MR^2, where M is the mass of the cylinder and R is the radius of the cylinder.

2. How does the moment of inertia of a rolling cylinder differ from that of a stationary cylinder?

The moment of inertia for a rolling cylinder is greater than that of a stationary cylinder due to the additional rotational motion. This means that more force is required to change the rotational motion of a rolling cylinder compared to a stationary one.

3. Does the shape of the cylinder affect its moment of inertia?

Yes, the shape of the cylinder does affect its moment of inertia. A solid cylinder will have a different moment of inertia than a hollow cylinder, even if they have the same mass and radius.

4. How can the moment of inertia for a rolling cylinder be used in real-world applications?

The moment of inertia for a rolling cylinder is important in engineering and physics applications, such as designing vehicles and analyzing their motion. It is also used in calculating the energy and power required for rotational motion.

5. Can the moment of inertia for a rolling cylinder be negative?

No, the moment of inertia for a rolling cylinder cannot be negative. It is always a positive value, as it represents the resistance to rotational motion.

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