Test 2 questions

Hopefully you've received my latest email, which indicates that the VCCS email system has been haywire all week but that they think the problem is resolved. Let's hope so.

Note that you will automatically get an extension if your work isn't completed by next Friday. With your Katrina-related absence early in the term, you've done an exceptional job getting to this point and appear to be on track to get finished by then, but physics takes time to sink in and give yourself the slack you need if it comes to that.

We used the SHM program on multiple computers both yesterday and today and encountered no problems. Are you using the program at Sup Study under Course Docs > Downloads > Physics I? If so let me know and I'll run the problem by our technical staff. I've never encountered that problem, but student do occasionally report such a message. A reboot usually fixes the problem.

See my notes and please follow up with additional questions.

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21:34:26 What will be the angular acceleration of the disk?

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RESPONSE --> Ahhh ok. m (kg) L (m) .007 .231 .024 .168 .047 .09 Force = .06893N rad = .3m Inertia = mr^2 = (.007kg * (.231m)^2) + (.024kg * (.168m)^2) + (.047kg * (.09m)^2) = .001432 kgm^2 *****TAU = I * ALPHA***** The relationship for rotational objects is angular acceleration = net torque / moment of inertia, in symbols alpha = tau_net / I. *****ALPHA = TAU-NET / I = .06893N / .001432kgm^2 = 48.14rad/s^2 ??*****

Tau represents torque, which is force * moment arm. You are using just force for the torque.

The moment arm here is the .30 meters from the axis of rotation to the rim of the disk.

In general the moment arm of a force is the distance from the axis of rotation to the line of application of the force. In this case the line of application is tangent to the circle, and the distance from the center of a circle to a point of tangency with the rim is equal to the radius.

See more about the use of torques in the Intro Problem Set on rotation.

As this example shows, and as you clearly know, it is essential to do the unit calculations and be sure they come out correctly.

Thinking about the units also helps build your understanding of the quantities are their relationships.

Your algebra skills are good so with practice you will master the details of unit calculations.

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21:35:00 ** The moment of inertia of the system is the sum of all the m r^2 contributions of the individual particles. The net torque is the product of the net force and the moment arm. Newton's 2d law F = m a, expressed in in angular form, is `tau = I `alpha, where `tau is net torque (analogous to force), I is moment of intertia (sum of all mr^2, analogous to mass) and `alpha is angular acceleration in rad/s^2. **

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RESPONSE --> Ok I did all that so i'm just hoping I used the right numbers...

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21:42:14 If the force is applied for 4 seconds with the disk initially at rest, what angular velocity with the disk attain?

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RESPONSE --> 'dt = 4s, v0 = 0, a = 48.14m/s^2 acceleration = 'dv / 'dt 'dv = a * 'dt = 48.14m/s^2 * 4s = 192.56m/s I think that's right...

This is a rotational situation and you need to work everything out in angular units.

The correspondence between angular units and linear units all follows from the definition of 1 radian as the angle for which the displacement along the arc is equal to the radius of the circle.

From this it follows that the distance `ds moved along the arc is equal to the product of the radius and the angle:

`ds = r * `dTheta (displacement along the arc is equal to the radius * the number of radians)

It follows by dividing both sides by `dt that `ds / `dt = r * `dTheta / `dt. Since v = `ds / `dt is the speed v of the point moving along the arc and omega = `dTheta / `dt is the angular velocity of the rotation, we have

v = r * omega (speed along the arc is equal to the radius * the number of radians per second).

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21:42:23 ** Once you know angular acceleration it's easy to find change in angular velocity **

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RESPONSE --> um ok... so did I do that right? You haven't calculated the angular acceleration. *****I thought I calculated the angular acceleration above and got ALPHA = TAU-NET / I = .06893N / .001432kgm^2 = 48.14rad/s^2***** Once you calculate the angular acceleration and multiply by `dt, you will get the angular velocity.

See my previous note on torque.

With regard to units:

A radian of angle multiplied by a meter of radius gives a meter of arc distance, so rad * m can equal m.

A meter of arc divided by a meter of radius gives radians of angle, so m / m can equal rad.

N / (kg m^2) would be (kg m/s^2) / (kg m^2), and the meter in the numerator could be divided by one of the meters in the denominator to give a radian of angle. However we would be left with rad / (s^2 * m), not rad/s^2.

Dividing torque by moment of inertia gives the units

m * N / (kg m^2) = m * (kg m/s^2) / (kg m^2) = rad / s^2,

where the rad comes from m / m and the other m / m just 'cancels out'.

Change in angular velocity = alpha * `dt.

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21:43:49 What then will be the speed of each of the masses? *****Omega = ‘ds from axis * ang. Velocity = .231m * 192.56rad/s = 44.5 (I’ve completely messed up my units here and don’t know if I’m using meters or radians/second or either of them at all.)

omega is angular velocity, so omega is not `ds * angular velocity.

v along the arc is r * omega, where r is the radius of the circular arc.

.231 m * 193 rad/s would give you 44.5 m * rad / s = 44.5 m / sec, where the m * rad is a meter of radius multiplied by a radian of angle and hence gives us a meter of arc.

The 44.5 represents the speed v along the arc of the circle, not an angular velocity, so you wouldn't multiply this by the .3 m radius.

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RESPONSE --> Yeah, I have no idea...

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21:48:45 ** You know angular velocity and distance of each mass from the axis of rotation. Angular velocity is the velocity of the mass along the arc divided by the radius. So what is the velocity of each of the masses along its arc? **

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RESPONSE --> m (kg) L (m) .007 .231 .024 .168 .047 .09 For some reason I've completely forgotten how to figure this out. v = r * omega. Velocity along the arc = distance from axis * angular velocity. *****Hopefully I did the right calculations for this above. I’m not really working well with the speed and velocity stuff… or any of this it seems. I hope there’s a curve for this class.*****

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21:50:15 What will be their total kinetic energy?

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RESPONSE --> KE = 1/2 I * omega^2 for each of the masses... since I couldn't figure out the velocity, I can't figure out the KE. This problem's just throwing me for a loop. *****KE = ˝ I * omega^2 = ˝ .001432kgm^2 * 94.18^2 = 6.35J I don’t know if this is right, but since it asked for the total KE, I used the total inertia and then added all of the ‘omega’s together to get the 94.18… I’m not sure if I need the inertia for each individual mass along with the omega for each one and then add them up, or if this method was ok.*****

If you multiply 1/2 I omega^2 you get angular KE.

If you calculate the speed of each mass (distance from axis * angular velocity, v = r * omega) you can get its 1/2 m v^2. If you do this for all the masses you get their total 1/2 m v^2, which is their total KE.

Both ways you get the same result, which is the point of this series of questions.

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21:50:20 *&*& Add up the individual kinetic energies.

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RESPONSE --> ok

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21:50:45 Compare the total kinetic energy to the change in the quantity .5 I `omega^2.

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RESPONSE --> I thought that was the equation I needed for the kinetic energy... ok I'm confused. Make the corrections I indicated above and send me a copy, along with any questions you still have. EXTRA TEST 2 QUESTIONS: ""A roller coaster is at the top of a circular loop of radius 14m. The roller coaster is turned upside down. How fast must it be moving for a a passenger sitting on a scale to register her usual weight?"" ***** I'm not sure where to begin with this one and haven't seen much like it in my notes. I found a section in Chapter 5 dealing with weightlessness but I'm not sure if that applies here or not. *****

The net force on an object moving in a circle at constant velocity v is m * centripetal accel = m v^2 / r. This is always the case, whenever something is moving at constant speed around a circle.

At the top of the loop the direction of the centripetal acceleration is toward the center of the circle, which from that point is downward.

The forces acting in the downward direction are the force m g of gravity and the force F exerted by the scale. If the scale registers her usual weight, then F = m g.

So the net force on the woman will be m g + F, or just m g + m g = 2 m g.

Therefore m v^2 / r = 2 m g.

This is an equation that can easily be solved for v in terms of r.

Since we're interested here in force rather than torque a direct approach to the force would be more straightforward. However you can get this result from angular considerations as follows:

The torque exerted by the gravitational force acts along a line which passes .239 m from the axis of rotation (which is the top of the pendulum).

The gravitational force is M g so the torque is M g * .239 m, where the uppercase M stands for the mass of the pendulum.

From this you could calculate the angular acceleration of the pendulum, but that isn't what we're after here.

The same torque could be obtained by a force perpendicular to the pendulum string. That force would be acting pretty much back toward the equilibrium position, and its line of action would be 3 meters from the axis of rotation at the top of the pendulum.

The force necessary would be the force F for which 3 meters * F = M g * .239 m. Solving for F we get

F = M g * .239 m / (3 m) = M * 9.8 m/s^2 * .239 / 3 = .68 kg * .79 m/s^2 = .54 N, approx..

The more conventional approach is as follows:

The displacement from equilibrium is small so the pendulum has very little vertical displacement or acceleration. So we very nearly have vertical equilibrium.

The result is that the vertical component of the tension is very nearly equal to the weight M g of the pendulum.

The tension vector is very close to vertical so its magnitude does not differ significantly from its vertical component.

So the tension is about equal to M g.

The horizontal component of the tension is M g sin(theta), where theta is the angle between the horizontal direction and the tension vector.

By similarity of triangles, we find that sin(theta) = .239 m / (3 m).

So the horizontal component of the tension is M g * .239 m / (3 m). This works out the same as the result obtained by considering torques.

The general rule is that for displacements x which are small compare to the length L, the force tending to pull a simple pendulum back to equilibrium is

F = - (M g / L) * x.

In the given instance M = .68 kg, x = .239 m and L = 3 m.