# Last 3 problems on Rotational Dynamics

1. Nov 8, 2007

### chavic

[SOLVED] Last 3 problems on Rotational Dynamics

OK, I've done problems 1-3, but I'm having a real hard time with 4-6. Any help you guys could give me would be greatly appreciated!

4. The problem statement, all variables and given/known data
A block of mass m1 = 2.45 kg and a block of mass m2 = 5.55 kg are connected by a massless string over a pulley in the shape of a solid disk having radius R = 0.250 m and mass M = 10.0 kg. These blocks are allowed to move on a fixed block-wedge of angle θ = 30.0°. The coefficient of kinetic friction is 0.360 for both blocks. Draw free-body diagrams of both blocks and of the pulley.

(a) Determine the acceleration of the two blocks.
(b) Determine the tensions in the string on both sides of the pulley.
Left:
Right:

4. Relevant equations
$$\sum$$$$\tau$$=RFsin$$\theta$$

net $$\tau$$=I$$\alpha$$

I=.5MR$$^{2}$$

$$\alpha$$= a/R

4. The attempt at a solution

I drew a Free Body Diagram and used the given information to find that:

• m$$_{1}$$=2.45 kg
m$$_{2}$$=5.55 kg
$$\mu$$=.36
R=.25 m
M=10 kg
$$\theta$$=30
n$$_{1}$$=24.01 N
n$$_{2}$$=47.1031
Force of Friction on M1=8.6436 N
Force of Friction on M2=16.9571 N
M1gsin$$\theta$$=27.195

• So some of forces acting on pulley is 1.5943 N to the right

• So $$\sum$$$$\tau$$=RFsin$$\theta$$

=.398575

• Plug this into net $$\tau$$=I$$\alpha$$
(I=.5MR$$^{2}$$)
(I=.3125)

.398575=.3125$$\alpha$$
1.27544=$$\alpha$$

• Plug this into $$\alpha$$= a/R
1.27544=a/.25
.31886=a

This is wrong and I tried several different times, went over my work and I can't find what I did wrong.

----------------------------------------------------------------------------------------​

5. The problem statement, all variables and given/known data
Two blocks, as shown below, are connected by a string of negligible mass passing over a pulley of radius 0.320 m and moment of inertia I. The block on the frictionless incline is moving up with a constant acceleration of 1.20 m/s2.

(a) Determine the tensions in the two parts of the string.
T1:
T2:
(b) Find the moment of inertia of the pulley.

5. Relevant equations
?

5. The attempt at a solution

----------------------------------------------------------------------------------------​

6. The problem statement, all variables and given/known data
The reel shown below has radius R and moment of inertia I. One end of the block of mass m is connected to a spring of force constant k, and the other end is fastened to a cord wrapped around the reel. The reel axle and the incline are frictionless. The reel is wound counterclockwise so that the spring stretches a distance d from its unstretched position and is then released from rest.

a) Find the angular speed of the reel when the spring is again unstretched. (Answer using theta for θ, g for the acceleration due to gravity, and R, I, m, k, and d, as necessary.)

(b) Evaluate the angular speed numerically at this point if I = 1.20 kg·m2, R = 0.300 m, k = 50.0 N/m, m = 0.500 kg, d = 0.200 m, and θ = 37.0°.

6. Relevant equations

?

6. The attempt at a solution

2. Nov 8, 2007

### hage567

I noted a couple of things:

You seem to have the right idea, but since you haven't really shown the equations you used to sum up the forces it isn't easy to see where you went wrong.

You should have 5 equations relating all of the variables. Can you show them? For example, for M1:

$$N_1 - m_1g = 0$$

$$T_1 - f_f = m_1a$$

Do the same thing for M2, leaving everything in terms of the variables for now.

The next thing to do is sum up the torques on the pulley, which you do according to your equations $$\Sigma \tau = RFsin\theta = I \alpha$$ (remember that the forces are acting tangentially on the pulley).

Once you've done this, you'll be able to find for a by solving the system of equations.

3. Nov 8, 2007

### hage567

For #5, You take the same approach as #4. This one is probably a bit easier than #4, since there is no friction involved. So sum up the forces using Newton's second law for each mass. You can then find the tensions because you have been given the acceleration of the system. To find I, sum up the torques and use the information from part (a).

For #6, you can do the same approach again.

Just be sure to draw careful diagrams, and label everything. This will help keep track of things.