# Ch 8.1 Help

## Ch 8.1 Help

__Question 1__A disk turns with an average angular velocity of 0.4173 rad/s. Through which angle in degrees does the disk turn in 2.613 sec (see sheet 3)? Indicate with a negative (positive) sign whether a point on the disk at larger distance r from the center of rotation moves faster (slower) than a point at smaller distance r from the center.

__How to Solve:__

(Answer is _______)

ω = (ΔΘ)/(Δt)

__Question 2__A point marked at the edge of a rotating disk which has a radius of 34.12 cm moves a distance of 9.562 m in 1.294 minutes. What is the average angular velocity of the disk (in SI units) (see sheet 4)? Indicate with a negative (positive) sign whether you can (cannot) use the same expression you used above to relate the instantaneous angular velocity and the instantaneous tangential velocity to each other.

__How to Solve:__

(Answer is Negative)

ω = v/r

To find our velocity just divide the seconds into the meters. Before we do that we have to convert those minutes into seconds. To do this multiply by 60.

time

_{seconds}= 1.294 minutes*60

time

_{seconds}= 77.64.

Now divide this into our meters

v = meters/seconds

v = 9.562/77.64

v = 0.1232

Convert your radius into meters

34.12cm*100 = 0.3412m

Now we can find the angular velocity (ω)

ω = v/r

ω = 0.1232/0.3412

ω =

- Spoiler:
- 0.3609559

__Question 3__A wheel with a 0.2772 meter radius, initially at rest, rolls down an incline and reaches after 5.885 seconds a linear velocity of 2.087 m/s. What is the angular acceleration of the wheel (see sheet 5 and the beginning of 34 where rolling is explained)? Indicate with a positive (negative) sign whether the tangential velocity of a point on the rim of the wheel is the same as (different from) the linear velocity of the wheel.

__How to Solve:__

(Answer is Positive)

α = a/r

Let's find our acceleration ([i]a[i]) whish is equal to velocity divided by the change in time (v/t). Just divide your velocity by the seconds that have passed

a = 2.087/5.885

a = 0.35463

Now we just divide by our radius,

α = 0.35463/0.2772

α =

- Spoiler:
- 1.27933

__Question 4__The constant tangential acceleration of a point at the edge of a rotating disk with a 28.2 m radius is 1.1898 m/s2. If the disk starts out with 21.58 rpm (see sheet 10 for "rpm") what is the angular velocity after 4 seconds (see sheet 5,7,10)? Indicate with a positive (negative) sign whether the initial angular velocity is (is not) given in SI units.

__How to Solve:__

(Answer is _______)

ΘΔπ√ωα

All I've really done so far.

**Guest01**- Posts : 133

Join date : 2008-09-19

## Re: Ch 8.1 Help

for number one im having problems, i know that's the equatrion but do we have to use tan-1 anywhere? how do u calculate the exact angle?

**thb1026**- Posts : 8

Join date : 2008-09-27

## Re: Ch 8.1 Help

You can look at angular motion (on a circle) and linear motion in a one-to-one correspondence. x->Θ, v->w, F->torque, et cetera. The units difference between angular motion and linear motion comes from what sort of "distance" you're measuring. For linear motion we use meter sticks, which gives us lengths of meters, and velocities of meters per second. We measure angles in terms of degrees or radians. A full circle has 360 degrees or two pi radians of curve, which is why the trigonometric functions are periodic on a two pi basis. from 360=2(pi) you can convert between radians and degrees at will.

The first question is really asking you to view the distance = (rate)(time) in view of angular motion, and then make a connection between the linear velocity and the angular velocity of a point on the disc. The relation v=(r)(w), where r is the distance of the point from the axis of rotation, should help you with the second part of the question.

The first question is really asking you to view the distance = (rate)(time) in view of angular motion, and then make a connection between the linear velocity and the angular velocity of a point on the disc. The relation v=(r)(w), where r is the distance of the point from the axis of rotation, should help you with the second part of the question.

**alogon**- Guest

## Re: Ch 8.1 Help

can someone please post the equation needed for #$ because the one posted previously is not clear as to what to do.

thanks

thanks

**Question**- Guest

## 8.1 Question #4

i dont really understand the equation used for question #4,can someone please repost the equation that i can used for question 4.

**Guest 0.**- Guest

## Questions 1 and 4

If anyone could please help with number 1 and number 4 I keep getting them wrong.

**sedwards**- Guest

## Re: Ch 8.1 Help

Question 1:

A disk turns with an average angular velocity of 0.8273 rad/s. Through which angle in degrees does the disk turn in 4.46 sec (see sheet 3)? Indicate with a negative (positive) sign whether a point on the disk at larger distance r from the center of rotation moves faster (slower) than a point at smaller distance r from the center.

How to Solve:

Answer is NEGATIVE

First, convert the radians/sec into degrees/sec

.8273 rad/s * (1/((2pi)/(360)) = 47.4008 deg/sec

Then you do 47.4008 deg/sec * 4.46 sec = 211.407

And that is your answer

Question 4:

The constant tangential acceleration of a point at the edge of a rotating disk with a 29.7 m radius is 0.956 m/s2. If the disk starts out with 22.25 rpm (see sheet 10 for "rpm") what is the angular velocity after 4 seconds (see sheet 5,7,10)? Indicate with a positive (negative) sign whether the initial angular velocity is (is not) given in SI units.

How to Solve:

Answer is NEGATIVE

angular acc = a/r

angular acc = .956/29.7 = .03218

22.25RPM * (2pi)/(60) = 2.330014

w = 2.330014 + .03218 (4) <- where 4 is time = 2.458 which is your answer

good luck!

A disk turns with an average angular velocity of 0.8273 rad/s. Through which angle in degrees does the disk turn in 4.46 sec (see sheet 3)? Indicate with a negative (positive) sign whether a point on the disk at larger distance r from the center of rotation moves faster (slower) than a point at smaller distance r from the center.

How to Solve:

Answer is NEGATIVE

First, convert the radians/sec into degrees/sec

.8273 rad/s * (1/((2pi)/(360)) = 47.4008 deg/sec

Then you do 47.4008 deg/sec * 4.46 sec = 211.407

And that is your answer

Question 4:

The constant tangential acceleration of a point at the edge of a rotating disk with a 29.7 m radius is 0.956 m/s2. If the disk starts out with 22.25 rpm (see sheet 10 for "rpm") what is the angular velocity after 4 seconds (see sheet 5,7,10)? Indicate with a positive (negative) sign whether the initial angular velocity is (is not) given in SI units.

How to Solve:

Answer is NEGATIVE

angular acc = a/r

angular acc = .956/29.7 = .03218

22.25RPM * (2pi)/(60) = 2.330014

w = 2.330014 + .03218 (4) <- where 4 is time = 2.458 which is your answer

good luck!

**Question**- Guest

## Does anyone Know how to figure out number 4 ??

I tried to figure out number 4 and labeled it positive and negative and am still getting the wrong answers and I am pretty sure I did the calculations correctly any insite?

THankS !

THankS !

**BGGirl**- Posts : 6

Join date : 2008-09-21

## Re: Ch 8.1 Help

Post your calculations, you both are probably messing up somewhere in converting wrong.

For question 1 it should be rad/s * 1/(2pi/360) [that is calculate 2*pi divided by 360, then divide that into 1.

For question 4, after you find your angular acc times that by 4. and then add that to your converted RPM. For the converted RPM it's 2*pi and divide that by 60 and take that and multiply it by your given RPM for your converted RPM

For question 1 it should be rad/s * 1/(2pi/360) [that is calculate 2*pi divided by 360, then divide that into 1.

For question 4, after you find your angular acc times that by 4. and then add that to your converted RPM. For the converted RPM it's 2*pi and divide that by 60 and take that and multiply it by your given RPM for your converted RPM

**Guest01**- Posts : 133

Join date : 2008-09-19

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