Rotational Kinematics/Momentum Problem

[jjd101]

Homework Statement

Flywheels are large massive wheels (disks) used to store energy. They can be spun up slowly, and then the wheel’s energy can be released quickly to accomplish a task that demands high power. A large flywheel has a 2.4 m diameter and a mass of 300 kg. Its maximum angular velocity is 1500 rpm.

a. The flywheel is spun up at a constant torque of 70 Nm. How long does it take the flywheel to reach top speed?

b. How much energy is stored in the flywheel?

c. After reaching top speed, the flywheel is connected to a machine to which it will deliver energy. Half of the energy stored in the flywheel is delivered in 2.5 sec. What is the average power delivered to the machine?

d. How much torque does the flywheel exert on the machine?

Homework Equations

I=1/2MR^2

The Attempt at a Solution

I calculated the moment for Inertia for this disk and got 216 but i am not sure what to do from here

 

[Redbelly98]

It's a good idea to check your textbook for the section or chapter that covers this topic. You will find many relevant formulas, and you can think about how they might apply to this problem.

Homework Statement

Flywheels are large massive wheels (disks) used to store energy. They can be spun up slowly, and then the wheel’s energy can be released quickly to accomplish a task that demands high power. A large flywheel has a 2.4 m diameter and a mass of 300 kg. Its maximum angular velocity is 1500 rpm.

a. The flywheel is spun up at a constant torque of 70 Nm. How long does it take the flywheel to reach top speed?

Check over all formulas that involve torque, and think about how one of them might apply here.

b. How much energy is stored in the flywheel?

Check for any formulas that relate energy and rotating objects.

c. After reaching top speed, the flywheel is connected to a machine to which it will deliver energy. Half of the energy stored in the flywheel is delivered in 2.5 sec. What is the average power delivered to the machine?

d. How much torque does the flywheel exert on the machine?

a. What is the amplitude of the subsequent oscillation?

b. What is the period of the subsequent oscillation?

Homework Equations

I=1/2MR^2

The Attempt at a Solution

I calculated the moment for Inertia for this disk and got 216 but i am not sure what to do from here

That's a start. As I said, check over your book discussion and look for formulas that involve whatever each question is asking for.

 

[netgypsy]

Notice your title uses rotational kinematics and momentum. Look up those formulas. yes you need moment of inertia but that's not enough.

 

[jjd101]

Krot = 1/2Iw^2 and T = mr^2(angular Acceleration) ?

 

[jjd101]

i solved for angular acceleration and got .162

 

[jjd101]

this doesn't seem right to me because i solved for time it takes to reach 1500 rpm(157.1 rad/sec) and got 970s

 

[netgypsy]

You have your torque is mrsquared alpha. Remember this is a disk, not a ring.

torque = I alpha so you need to use the correct moment of inertia.

Once you get angular acceleration you can use angular kinematics equations and rotational energy equations to calculate the other variables you need.

 

[jjd101]

i did this and solved for alpha with the correct moment of inertia and alpha was .324 giving 484.9s as time to reach 1500 rpms. for Krot=1/2Iw^2 do i use w as the 1500 rpm in rad/sec?

 

[netgypsy]

wait - alpha is in rad/secsquared yes omega is in rad/sec

 

[jjd101]

so Krot= 1/2I(omega)^2 Krot= (1/2)(216)(157.1)^2 Krot= 2665484.3? huh?

 

[netgypsy]

Flywheels store a lot of energy. If your calculation is right, your answer could be huge. You have waaay too many significant digits though. Your given info has only two so round it and put it in exponential notation.

 

[jjd101]

2.7*10^6 and with this i got the answer to c as 540,000 Watts. how do i solve for d?

 

[netgypsy]

work is torque theta and since power is time rate of change of work, what will it be equal to in terms of torque?

 

[jjd101]

T= (P*(delta t))/Theta?

 

[netgypsy]

time rate of change of work means Work divided by time. so what happens to the other side of the equation if you divide one side by time?

 

[jjd101]

wouldnt you multiply by T since W/T= P to get W=PT I am confused as to what youre saying

 

[netgypsy]

You start with the fact that work is difined as torque times the angle for rotational motion.

But you don't have the angle the machine rotates through so you can't calculate the torque from this. BUT you know that power is defined as work divided by time so take what work is equal to in terms of torque and divide it by time because you also have power.

 

[jjd101]

So W = T(theta)/Time? What is theta in this problem?

 

[netgypsy]

ahh you don't know theta but what variable is theta/time?

 

[jjd101]

theta/time is omega? so W=Tomega W/omega=T T=540000/157.1 = 3437.3 Nm?

 

[netgypsy]

not work but POWER = Torque omega and you have power and I thought you had omega also after the energy was transferred.

 

[jjd101]

sorry i meant power, 540000 is the value for power, isn't omega 1500 rpm converted to rad/sec which is 157.1rad/sec?

 

[netgypsy]

It seems like it has to be. This part bothers me a bit and I'm brain dead. What bothers me is we don't know anything about the machine and omega isn't necessarily the same for the machine and the fly wheel. The linear velocity is the same where they are attached to a belt but not necessarily the angular velocity. If you have the answers try it. I'll check it out again tomorrow and see if I can use linear values to double check or perhaps someone less brain dead will chime in.

 

[jjd101]

ok thank you