Work and Rotational Kinetic Energy

[G-reg]

Homework Statement

(a) If R = 12 cm, M = 360 g, and m = 20 g (below), find the speed of the block after it has descended 50 cm starting from rest. Solve the problem using energy conservation principles. (Treat the pulley as a uniform disk.)

Homework Equations

KEi + PEi = KEf + PEf

The Attempt at a Solution

0 + mgh = (1/2)mv^2 + (1/2)Iw^2 + 0

mgh = (1/2)mv^2 + [(1/2)(1/2)mr^2](v/r)^2

mgh = (1/2)mv^2 + [(1/4)mr^2](v/r)^2

4mgh = 2mv^2 + mr^2(v/r)^2

but I don't this has got to be wrong somewhere because the r's will cancel out and I know that they need to be in the equation for velocity..can anyone help? I would greatly appreciate it!

 

[anathema]

Thank you for posting your question. I am a scientist and I would be happy to assist you with solving this problem using energy conservation principles.

First, let's review the given information. We have a block with mass 360 g attached to a string that is wrapped around a pulley with radius 12 cm. The other end of the string is attached to a smaller block with mass 20 g, which is hanging below the pulley. The block starts from rest and descends 50 cm.

To solve this problem, we will use the equation for conservation of energy, which states that the initial energy of the system is equal to the final energy of the system. In this case, the initial energy is the potential energy of the 360 g block at a height of 50 cm, and the final energy is the kinetic energy of both blocks.

Let's start by calculating the initial potential energy of the 360 g block. We can use the formula PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height. Plugging in the given values, we get:

PEi = (0.360 kg)(9.8 m/s^2)(0.50 m) = 1.764 J

Now, let's consider the final energy of the system. The system consists of both blocks, so we need to calculate the kinetic energy of both blocks and add them together. The formula for kinetic energy is KE = 1/2 mv^2, where m is the mass and v is the velocity.

For the 360 g block, we can use the formula KE = 1/2 mv^2, where m is the mass and v is the velocity. The block starts from rest, so its initial velocity is 0. The final velocity will be the same for both blocks, since they are connected by a string and move together. Therefore, the kinetic energy for the 360 g block is:

KEf = (1/2)(0.360 kg)(v^2) = 0.180 v^2 J

For the 20 g block, we can use the same formula KE = 1/2 mv^2, where m is the mass and v is the velocity. Again, the block starts from rest, so its initial velocity is 0. The final velocity will be the same as the 360 g block, so the kinetic

 

[tomcue]

Your attempt at a solution is on the right track, but there are a few errors in your equations. First, the rotational kinetic energy term should be (1/2)Iω^2, where I is the moment of inertia of the pulley. For a uniform disk, the moment of inertia is (1/2)mr^2. Also, you did not include the mass of the pulley in your equations, which will affect the final velocity. The correct equation should be:

mgh = (1/2)mv^2 + (1/2)Iω^2

Substituting in the moment of inertia for a uniform disk, (1/2)mr^2, and the angular velocity, v/r, we get:

mgh = (1/2)mv^2 + (1/2)(1/2)mr^2(v/r)^2

Simplifying, we get:

mgh = (1/2)mv^2 + (1/4)mv^2

Solving for v, we get:

v = √(8gh/3)

Substituting in the given values, we get:

v = √(8*9.8*0.5/3) = 2.42 m/s

So the final speed of the block after descending 50 cm is 2.42 m/s. This solution assumes that there is no friction in the system. If there is friction, it will cause a decrease in the final velocity.