Rotational Motion Two Questions

[hunansux]

Homework Statement

A 16.0 kg bucket of water of is suspended by a rope wrapped around a pulley, which is just a solid cylinder 0.300 m in diameter with mass 12.0 kg. The cylinder is pivoted on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls 12.0 m to the water. Neglect the weight of the rope.

Homework Equations

Conservation of Energy, Rotational Kinematics Equations

The Attempt at a Solution

Solving for tension in the rope (wrong):

Ft - Fg = ma

Speed of the bucket as it hits the water (also wrong..):

mgyi + (mvi^2)/2 = mgyf + (mvf^2)/2

While the bucket is falling, what is the force exerted on the cylinder by the axle?

^^ No idea how to do that ****.



Problem 2

A 15.0 kg mass is attached to a cord that is wrapped around a wheel of radius r = 9.0 cm (Fig. P8.60). The acceleration of the mass down the frictionless incline is measured to be 2.00 m/s2. Assume the axle of the wheel to be frictionless.

Questions:
Determine the tension in the rope.
Determine the moment of inertia of the wheel.
Determine the angular speed of the wheel 2.00 s after it begins rotating, starting from rest.

This is really similar to the previous one and since I couldn't figure that out... I thought I'd see if you guys could help me.

 

[pabloenigma]

for the second question u haven't mentioned the angle of incline.if the angle of incline is y. then we have the force equation as

mgsiny-T=m x 2.00m/s2.

also from the rotation of the wheel,
2.00m/s2=alpha x radius of wheel.(gives u the angular acc)

now
tau=I x alpha gives u the MI,

and find out the angular velocity from omega=alpha x t.
...

 

[hunansux]

Whoops my bad the angle is 37 degrees it was in the picture not the prompt though! Sorry

Edit: "tau= I x alpha" How do I find the torque and the alpha? Sorry for the dumb question I'm sure it has an incredible easy solution... I just got moved in a AP Physics class from Honors so I'm missing a lot.

 

[pabloenigma]

Whoops my bad the angle is 37 degrees it was in the picture not the prompt though! Sorry

Maybe there's a problem with my pc,but I don't see any pic,nor any links!

 

[hunansux]

The picture is from my textbook haha I didn't mean online! Thanks for the help though. I got the first part of the second question

 

[pabloenigma]

Whoops my bad the angle is 37 degrees it was in the picture not the prompt though! Sorry

Edit: "tau= I x alpha" How do I find the torque and the alpha? S

tau is the product of the tension and the radius of cylinder...
alpha is obtained from the previous eqn(see my original soln)

 

[hunansux]

haha I guess I should have googled tau I assumed it was another way of saying torque... anyway thanks for the help! I still need help one number one. No idea what I did wrong I followed the example we did in class... might have been sleeping though haha. Anyway thanks!

 

[tiny-tim]

welcome to pf!

hi hunansux! welcome to pf!

(btw, not a good idea to put two questions in the same thread)

A 16.0 kg bucket of water of is suspended by a rope wrapped around a pulley, which is just a solid cylinder 0.300 m in diameter with mass 12.0 kg. The cylinder is pivoted on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls 12.0 m to the water. Neglect the weight of the rope.
…
Solving for tension in the rope (wrong):

Ft - Fg = ma

Speed of the bucket as it hits the water (also wrong..):

mgyi + (mvi^2)/2 = mgyf + (mvf^2)/2

While the bucket is falling, what is the force exerted on the cylinder by the axle?

try again, using the kinetic energy or angular momentum of the pulley

 

[hunansux]

Thanks for the help I'll try that too. I sent my teacher an email no response though :( Would it have been better to make two threads?

 

[SammyS]

Homework Statement

A 16.0 kg bucket of water of is suspended by a rope wrapped around a pulley, which is just a solid cylinder 0.300 m in diameter with mass 12.0 kg. The cylinder is pivoted on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls 12.0 m to the water. Neglect the weight of the rope.

Homework Equations

Conservation of Energy, Rotational Kinematics Equations

The Attempt at a Solution

Solving for tension in the rope (wrong):

Ft - Fg = ma

Speed of the bucket as it hits the water (also wrong..):

mgyi + (mvi^2)/2 = mgyf + (mvf^2)/2

While the bucket is falling, what is the force exerted on the cylinder by the axle?

^^ No idea how to do that ****.

You can solve this by finding the acceleration, a, of the bucket, and then by use kinematics.

Alternatively, you can use conservation of energy.

Let's look at using conservation of energy.

You had written: mgy_i + (mv_i²)/2 = mgy_f + (mv_f²)/2.

This neglects the kinetic (rotational) energy of the pulley.

The Kinetic Energy of a rotating object is: (1/2)I(ω²), where I is the object's moment of inertia, and ω is the object's angular velocity (in radians/second).

So you should have written something like:

mgy_i + (½)m(v_i)² + (½)I(ω_i)² = mgy_f + (½)m(v_f)² + (½)I(ω_f)².

Of course, v_i, ω_i, and y_f are all zero.

You need to find the moment of inertia of a solid cylinder. Also relate the speed of the bucket to the angular velocity of the pulley.

 

[hunansux]

Thanks for the help sammy. Any idea on how to do that third part? I'm struggling with that.

 

[SammyS]

Thanks for the help sammy. Any idea on how to do that third part? I'm struggling with that.

The third part ... Do you mean, "While the bucket is falling, what is the force exerted on the cylinder by the axle? " ?

The linear acceleration of the cylinder is zero, so the net force on the cylinder is zero.

The downward forces on the cylinder are:

1. The tension in the rope.

2. The weight of the cylinder = mg of the cylinder.​

The upward force on the cylinder is provided by the axle, and is equal (but opposite) to the sum of the two downward forces.

Next question to be asked is: "What is the tension in the rope?" No, it's not equal to the weight of the bucket. If it were, the bucket wouldn't accelerate downward!

T - (m_b)*g = (m_b)*a, where T is the tension, m_b is the mass of the bucket, a is the acceleration of the bucket (assumed to be negative, the way I wrote it).

Do you know the acceleration?

Spoiler

v_f² - v_i² = 2*(a)*(y_f - y_i)

 

[hunansux]

Thanks everyone for the help got a chance to talk to my teach today. He helped with the rest! Admins can close this if they'd like to!