# Linear angular wheel inertia mass problem

• greg997
In summary, we have a wheel with a moment of inertia I and a mass M=14 kg and a radius of gyration k=0,15 m. A cord is wrapped around the drum with a radius R= 0,09 m and a mass m=0,1 kg is hung on the end. The mass is allowed to fall a distance s= 2,5 m before it hits the ground. The acceleration of the mass a m/s^2 is uniform. The corresponding angular acceleration of the wheel is alfa rad/s^2. The mass exerts a force on the cord of m(g-a). The resulting angular acceleration is given by the formula: alfa= mg/((I/R)+mR

#### greg997

We've got a wheel with a moment of inertia I. The wheel has a mass M=14 kg and a radius of gyration k=0,15 m. A cord is wrapped around the drum with a radius R= 0,09 m and a mass m=0,1 kg is hung on the end. The mass i allowed to fall a distance s= 2,5 m before it hits the ground. The acceleration of the mass a m/s^2 is uniform. The coresponding angular acceleration of the wheel is alfa rad/s^2. The mass exerts a force on the cord of m(g-a). The resulting angular acceleration is given by the following formula : alfa= mg/((I/R)+mR).

1.Calculate the angular acceleration of the wheel alfa

I calculated the I=Mk^2
I= 14x0,15^2= 0,315kgm^2

Then angular acceleration from the formula above alfa= 0,2795 rad/s^2

Is that right?

2. calculate linear acceleration of the mass
3. velocity of the mass just before it hits the ground
4. angular velocity of the wheel just before the mass hits the ground.

Could anyone help mi with this? Thanks

Hi greg997,

For #2: Since the falling mass is attached directly to the outside of the drum by the rope, the linear acceleration of the mass is the same as the linear acceleration of a point on the surface of the drum. Now that you have the angular acceleration, you can find the linear acceleration of any point on the drum's surface (since you also know its radius). What do you get?

That should also help with #3 and #4.

hi alphysicist. Thank you for helping me.
2. i calulated the linear acceleration a= R alpha ; a= 0,025m/s2
3. velocity of the mass before it hits the ground i used formula v^2= u^2+2as what gives me v=1,118m/s
4. angular velocity of the wheel before the mass hits the ground i used formula v=Rw, w=v/R

Is that right? Thanks again for help

greg997 said:
hi alphysicist. Thank you for helping me.
2. i calulated the linear acceleration a= R alpha ; a= 0,025m/s2
3. velocity of the mass before it hits the ground i used formula v^2= u^2+2as what gives me v=1,118m/s

I think this may be incorrect. What numbers were you using to get the velocity?

:( I thin i know where i made error.
so the V should be v= 0,353m/s and the angular velocity of the drum w= 3.928 rad/s
Thanks for noticing that error in calculations.

I ve got next problem with the energy..
1. potential energy in the system before the mass falls.
I've used formula PE=mgz, where z= height=2,5m, m= M+m=14kg+0,1kg
PE= 345,8025.
2. kinetic energy of the falling mass just before it hits the ground
KE=(mV^2)/2 , m= 0,1 v=0,353
KE=0,00623045
3. kinetic energy of the wheel just before the mass hits the ground
KE=(Iw^2)/2 , I=0,315, w=3,928
KE=2,43 jouls
Is that corect?

4. Iam supposed to show that initial potential energy is equal to the final kinetic energy. how? the sum of KE of the wheel and KE of the mass doesn't equal PE.

Last edited:
You don't include the wheel's mass in the potential energy formula; what is important is the change in height for each mass, and the height of the wheel is not changing.

Remember that where their statement is coming from is from saying that energy is conserved. If your energy is potential and kinetic for both wheel and the smaller mass, then the potential energy of the wheel would always be the same number on both sides of the equation and would therefore cancel.

Thanks again for your help. Where are you? USA? I am in the Uk. But i come from Poland.
Best wishes

I'm from the USA. That's why I also had a brief moment of confusion when you said the velocity was 1,118 m/s. I'm used to that meaning over 1 km/s!

Hi, could you help me wit one more question?
Could you explain to me what affect friction would have on the predicted velocities. We have that wheel, and mass hang on a cord. I would say that the predicted velocities would be smaller or even there would be no motion at all.
Thank you

That sounds right to me.

Hi, its me again. Coud you help me again. I am not sure how the graphs should look like.
I have a mass of 4kg suspended from a spring and oscillating up and down at 2Hz.
The amplitude is 5mm.;determione displacement, velocity, acceleration 0.02 s after the mass passes through the rest position in an upwards direction.
I calculated that stiffness k=631.01 N/m
displacement x= 1.24 mm
velocity v= 60.82
acceleration a= -196.05

how to draw these graphs? does the sine wave starts at 0?

That's looks right (of course your acceleration is in mm/s^2; is that what you wanted?).

I think the easiest way is to just draw a sine curve, and then label the following points: on the vertical axis, the amplitudes of the wave; on the horizontal axis, the times when the particle passes through the point x=0, and maybe those times when the mass is at the amplitudes.

I think I got it right now. Thanks again for helping me. I am really gratful.

so a vehicle has a mass = 1000kg and is traveling around a circular bend of radius 140 m. The centripetal force on the vehicle = 1100N.
1. calculate the velocity in km/h

I got 44.64
2. the acceleration in m/s2

I got 1.0982

3. the amount of g acting on the vehicle

and this is where I am stuck.

how to calculate this?

Thanks. I am really grateful.

Hi greg997,

Was that the exact wording of the question for part 3? I don't really understand what it's asking for.

you see, it is exact wording:(I ve got no clue.

If that's the exact wording, then "the amount of g acting on the vehicle" doesn't make sense to me.

If it had asked "the total acceleration in terms of g acting on the vehicle" or something like that, that would make sense, but that doesn't seem to be what it's asking for.

hi, i found out that it was just about dividing acceleration of the car a by g. That gives about 0.11. It's useless. Thanks again for helping me.