Ch 8.1 Help
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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 =
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 =
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.
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 : 20080919
Re: Ch 8.1 Help
for number one im having problems, i know that's the equatrion but do we have to use tan1 anywhere? how do u calculate the exact angle?
thb1026 Posts : 8
Join date : 20080927
Re: Ch 8.1 Help
You can look at angular motion (on a circle) and linear motion in a onetoone 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 : 20080921
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 : 20080919
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