Some problems with rotational motion

You have the correct value for 'r' in meters and 'I' and 'alpha' are correct. Also, I think you are right about #3.So...In summary, we discussed the calculations for the moment of inertia and torque needed to keep a ball rotating at a constant angular velocity, the required steady force of tangential rockets to correctly spin a satellite, and the total kinetic energy of a rolling bowling ball. We also corrected an error in one of the calculations and clarified the units used for torque.
  • #1
jrd007
159
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1) A .65 kg on the end of thin light rod is rotated in a horizontial circle of a radius 1.2 m. (a) Calculate the moment of inertia of the ball about the center of the circle, and (b) the torgue needed to keep the ball rotating at constant angular velocity if air resistence exerts a force of .020 N on the ball. (ignore rod's moment of I and air resistence).

So... I did find (a) successfully: I = mr^2 = .94 kg-m^2

And here were my thoughs for part (b): T = F(perpindic)r
T = I*a(angular) Since acceleration is constant is it not zero? correct answer: .0240 mN
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2)To get a flat uniform cylindrical satellite spinning at the correct rate, engineers fire four tangential rockets. If the satellite has a mass of 3600 kg and a radius of 4 m, what is the require steady force of each rockets if the satellite is to reach 32 rpm in 5 minutes?

This one really got me thinking. So first converted all I needed to. I was give:
m = 3600 kg, r = 4 m, w = 3.349 rads/2, = 300 s.

I know I have to use the F = ma, and I have the m so all I need to do is some how get the a, and that is where I am having trouble. Correct answer: 20 N
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3) A bowling ball of mass 7.3 kg and radius 9 cm = .09 m, rolls without slipping down a lane at 3.3 m/s. Calculate it's total kinetic energy.

What I did was use the equation: (1/2)Iw^2

First I found I by I=mr^2
I = .059

then I found w by w = v/r = 3.3/.09
w = 36.6

so .5(.059)(36.6^2)
= 39.5 J and the correct answer is 56 J What did I do wrong?
 
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  • #2
for problem 3,

one needs to find the translational and rotational KE's.

Obviously, KE (trans) = 1/2 mv2.

Rotational KE = 1/2 I [itex]\omega[/itex]2.

Make sure one has proper moment of inertia - in this case for a sphere I = 2/5 mR2, and when one multiplies R2 by [itex]\omega[/itex]2, the product is just v2.

For problem 2, try torque T = I[itex]\,\alpha[/itex], and [itex]\omega[/itex] = [itex]\omega_o[/itex] + [itex]\alpha\,[/itex]t, where [itex]\alpha[/itex] is angular acceleration and [itex]\omega[/itex] is angular velocity, and [itex]\omega_o[/itex] is the initial angular velocity.

For problem 1,
And here were my thoughs for part (b): T = F(perpindic)r
T = I*a(angular) Since acceleration is constant is it not zero?
To keep the ball moving, the applied torque must equal the opposing torque, which you have stated as T = F(perpindic)r. Simply subsitute in the values given. I think the units given with the correct answer are not correct, but should be N, not mN.
 
  • #3
Problem three:
.5mv^2 = (.5)(7.3 kg)(3.3 m/s)^2 = 39.7 J

.5Iw^2 = finding I = 2/5(7.3 kg)(.09 m)^2 = .025

so... .5(.025)(w) and w = v/r = (3.3 m/s)/(.09) = 37
so (.5)(.025)(36.6) = 16.7

adding them together is 16.7 J + 39.7 J = 56.4 = 56 J. Would that be correct?
-----------

#2 I used w = w(i) + a(ang)t
3.349 rad/s = 0 + a(300 s )
ang. acc = .011 m/s^2

then I found I, which fora cylinder is 1/2mr^2
= .5(3600 kg)(4 m)^2 ---- I = 28800

then I substitute in T = (28800)(.011 m/s^2) = 316... still not the right answer, why is that?
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#1, T = F(perpindic)r = (.020 N)(1.2 m) = .024 Nm, that is correct! At least I got two of them understood. Thanks. Any idea on # 2? Also, why are we only considering the force the air resistance is creating for #1 and now the mass?
 
  • #4
Try this on #2,

One is trying to calculate the Torque provided by 4 rocket motors (jets) at radius 'r', tangential to the circumference of the satellite (cylinder).

So torque T = 4 F x r, which is the total force at a moment arm of length r. Also the torque = Moment of Inertia x angular acceleration, which you have calculated correctly.

Equate 4 F r = I [itex]\alpha[/itex] and solve for F.


On problem #1, I misread your units. I was thinking mN meant milliNewtons, rather than N-m (Newton-meters), which is the correct unit for torque.
 
  • #5
Okay I shall try again.
4 F r = I [itex]\alpha[/itex]

4(F)(4 m) = (28800)(.011 m/s^2)
= 19.8 = 20 N, did I do the equation correctly?
 
  • #6
That seems to be correct.
 

1. What is rotational motion?

Rotational motion is the movement of an object around a fixed axis. This type of motion can be seen in objects such as a spinning top, a rotating wheel, or a swinging pendulum.

2. What are some common problems with rotational motion?

Some common problems with rotational motion include understanding the relationship between linear and rotational motion, calculating rotational speed and acceleration, and determining the axis of rotation.

3. How does rotational motion differ from linear motion?

Rotational motion involves movement around a fixed axis, while linear motion involves movement in a straight line. Additionally, rotational motion is described using angular quantities such as angular velocity and angular acceleration, while linear motion is described using linear quantities such as velocity and acceleration.

4. How can we calculate rotational speed and acceleration?

To calculate rotational speed, we use the equation ω = Δθ/Δt, where ω represents angular velocity, Δθ represents the change in angle, and Δt represents the change in time. To calculate rotational acceleration, we use the equation α = Δω/Δt, where α represents angular acceleration and Δω represents the change in angular velocity.

5. Why is understanding rotational motion important?

Understanding rotational motion is important because it allows us to explain and predict the behavior of objects in motion. It is also essential in many fields of science and engineering, such as mechanics, robotics, and astronomy.

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