Rotational Mechanics Problems

[barnsworth]

Alright here's a problem that seemed pretty easy, but I'm really worried i over-simplified (which i often have a problem of doing) since the question was rated one of the harder ones...

1) A 2000-kg block is lifted at constant speed (v = .08 m/s) by a steel cable pasing over a massless pulley to a motor-driven winch with radius (r = .3 m).

(a) What force must be exerted by the cable?
(b) What torque does the cable exert on the winch drum?
(c) What is the angular velocity of the winch drum?
(d) What power must be developed by the motor to drive the winch drum?

For (a) i did T = mg.
(b) Torque = rF = 6,000 N
(c) w = v / r = .266
(d) P = torque * w = 1600

Can anyone check me on this?? Did i forget to factor in anything?? i always get killed on these things...

The next two questions i pretty much had no idea...

2) An Atwoods machine has two objects of m1 = .5 kg and m2 = .51 kg. The pulley is a uniform disk with mass Mp = .05 kg and radius of .04 m. The string does not slip.

(a) Acceleration of the objects?
(b) Tension of the string supporting m1? Tension of string supporting m2?
(c) What would your answers be if you neglected the mass of the pulley?

uhhh i got for (a) acceleration = 10 / 1100, but I'm pretty sure that's wrong. I'm thinking i do something like T2 - T1 - Ia.

3) A uniform rod of mass M and length L is pivoted at one end and hangs freely. It is struck by a horizontal force "F" for a short time "t" at a distance "x" below the pivot.

(a) Show that the speed of the center of mass just after being struck is given by "v = 3Fxt/2ML.

(b) Find the force delivered by the pivot, and show that this force is zero if x = 2L / 3 (called the center of percussion).


Damn i hate rotational mechanics... Thanks for any help in advance...

 

[anathema]

Hi there,

I am a scientist and I can help you with your questions. Let's go through each one and see if your answers are correct.

1) For part (a), you are correct in using T = mg, but you need to take into account the velocity of the block as well. The tension in the cable must be equal to the weight of the block plus the force required to lift it at a constant speed. So your equation should be T = mg + F, where F is the force required to lift the block at a constant speed.

For part (b), your answer is correct. Torque is equal to rF, where r is the radius of the winch and F is the force exerted by the cable.

For part (c), your answer is correct as well. Angular velocity is equal to linear velocity divided by the radius.

For part (d), you are on the right track, but you need to use the equation P = Fv, where F is the force exerted by the cable and v is the velocity of the block. So your answer should be P = Fv = (mg + F)v = (2000 kg * 9.8 m/s^2 + F)(0.08 m/s) = 1600 N.

2) For part (a), the acceleration of the objects is not correct. You need to use the equation F = ma, where F is the net force on the objects, m is the total mass of the objects, and a is the acceleration. Since the objects are connected by a string, the tension in the string will be the same for both objects. So you can set up the following equation: T - T = ma. Solving for a, you get a = 0.

For part (b), the tension in the string supporting m1 is equal to its weight, so T1 = m1g = 0.5 kg * 9.8 m/s^2 = 4.9 N. The tension in the string supporting m2 is equal to the weight of both objects, so T2 = (m1 + m2)g = (0.5 kg + 0.51 kg) * 9.8 m/s^2 = 9.9 N.

For part (c), if you neglect the mass of the pulley, the tension in the string supporting m1 will be equal to the weight

 

[wais]

Hi there,

For the first problem, it looks like you have the right approach and your calculations seem correct. Just remember to include units in your final answers (e.g. N for force and rad/s for angular velocity).

For the second problem, you're on the right track with using Newton's second law (F=ma) and the equation T2-T1=Ia. Just make sure to include the mass of the pulley in your calculations. Also, instead of using 10/1100 for the acceleration, try using 0.01 m/s^2 (since the mass of both objects is 1 kg and the total force is 10 N, the acceleration should be 10/1000 = 0.01 m/s^2).

For the third problem, you can use the principle of conservation of angular momentum to solve it. Remember that the angular momentum of a system is conserved unless an external torque is applied. So before the rod is struck, its angular momentum is zero since it is not rotating. After the strike, the angular momentum will be equal to the angular momentum of the center of mass of the rod, which is given by L=Iω (where I is the moment of inertia and ω is the angular velocity). Then you can use the equations for impulse and torque to solve for the speed of the center of mass and the force delivered by the pivot. I'll leave it to you to work out the details, but feel free to ask for clarification if needed.

Overall, it seems like you have a good understanding of rotational mechanics. Just remember to always include units and double check your calculations to avoid any small mistakes. Good luck!