Question regarding block and pulley system with inertia

In summary, the homework equations are correct, but the third term in the middle parentheses is (I/r)^2...how is this possible? Shouldn't it be I/r^2? Any input would be greatly appreciated! Thanks!
  • #1
jcruise322
36
1
This problem is an example from Mosca and Tipler, 9-13, 6th edition. I believe the books equation for the acceleration of the system is incorrect according to my work... Anyway, here goes

1. Homework Statement

"Two blocks are connected by a string that passes over a disk pulley of radius R and moment of inertia I. The block of mass m1 slides on a frictionless, horizontal surface; the block of mass M2 is suspended from a string over the pulley. Find the acceleration of the blocks and the tensions. The string does not slip on the pulley."

Homework Equations


Non slip, so: a=r*α
For pulley: r(T2-T2)=I*α
For block 2: m2g-T2=m2*a m2=mass two
For block 1: T1=m1*a m1=mass one

The Attempt at a Solution


[/B]
Well, for the pulley: T2-T1=(I*α)r, and therefore: T2-T1=(I*a)/r^2 (substituting a/r=α).

I added the above equation to the two others so the tensions would cancel:

so: m2*g=(m2+m1+I/r^2)*a

so: a=(m2/(m2+m1+I/r^2))*g
The books says the the third term in the middle parentheses is (I/r)^2...how is this possible? Shouldn't it be I/r^2? Any input would be greatly appreciated! Thanks! :)
 
Physics news on Phys.org
  • #2
Check the dimensions - you can only add a mass to m1 and m2, so a term ##I / r^2## is acceptable, a term ##(I/r)^2## is surely not.
 
  • Like
Likes jcruise322
  • #3
jcruise322 said:
Non slip, so: a=r*α
For pulley: r(T2-T2)=I*α
For block 2: m2g-T2=m2*a m2=mass two
For block 1: T1=m1*a m1=mass one

Everything looks correct.

jcruise322 said:
Well, for the pulley: T2-T1=(I*α)r, and therefore: T2-T1=(I*a)/r^2 (substituting a/r=α).

I added the above equation to the two others so the tensions would cancel:

so: m2*g=(m2+m1+I/r^2)*a

so: a=(m2/(m2+m1+I/r^2))*g

You're right, from the torques acting on the pulley you get:

T2-T1=[(Mass of the pulley)*R*α]/2 = (I*α)/R

and from the no slip condition you know that α=a/R therefore

T2-T1=[(Mass of the pulley)*a]/2 = (I*a) / R^2

and solving for a, you get:
a=(m2/(m2+m1+I/R^2))*g, which is what you did.
 
  • Like
Likes jcruise322
  • #4
BvU said:
Check the dimensions - you can only add a mass to m1 and m2, so a term I/r2I / r^2 is acceptable, a term (I/r)2(I/r)^2 is surely not.
BvU is right!
 
  • Like
Likes jcruise322 and BvU
  • #5
Awesome, thanks guys! I will send Mosca and Tipler a "tip" on how they can rewrite their physics book haha. We use it for engineering here at UW Seattle :) Again, really appreciated :) :)
 
  • Like
Likes (Ron)^2=-1

Related to Question regarding block and pulley system with inertia

1. What is a block and pulley system?

A block and pulley system is a mechanical device that consists of a block (a stationary object) and a pulley (a wheel with a grooved rim) connected by a rope or cable. This system is used to change the direction of a force or to multiply the force applied.

2. How does inertia affect a block and pulley system?

Inertia is the tendency of an object to resist changes in its state of motion. In a block and pulley system, inertia affects the movement of the block and the pulley. The greater the inertia of the block, the more force is needed to move it. Similarly, the greater the inertia of the pulley, the more force is needed to change its rotational motion.

3. What is the purpose of using a block and pulley system with inertia?

The main purpose of using a block and pulley system with inertia is to reduce the amount of force needed to move an object. By increasing the inertia of the pulley, the force needed to lift the block is reduced as the pulley acts as a lever, multiplying the force applied.

4. How can the inertia of a pulley be increased in a block and pulley system?

The inertia of a pulley can be increased by increasing its mass or the distance of its mass from the axis of rotation. This can be achieved by adding weights to the pulley or increasing the size of the pulley's rim.

5. Are there any real-life applications of a block and pulley system with inertia?

Yes, there are many real-life applications of a block and pulley system with inertia. Some examples include cranes, elevators, and construction machinery. These systems use pulleys with increased inertia to lift heavy loads with less force.

Similar threads

  • Introductory Physics Homework Help
3
Replies
102
Views
5K
  • Introductory Physics Homework Help
Replies
3
Views
884
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
4K
  • Introductory Physics Homework Help
Replies
34
Views
2K
  • Introductory Physics Homework Help
Replies
14
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
17
Views
1K
  • Introductory Physics Homework Help
Replies
26
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
1K
Back
Top