How to find angle from angular velocity?

[Sneakatone]

A cat swipes at a spool of thread, which then rolls across thefloor with an initial speed of 1.5 m/s. The spool deceleratesuniformly to a stop 3 m from its initial position. The spool has aradius of 1.5 cm and rolls without slipping.


a) What is the initial angular velocity?
1.5 /(1.5/100)=100 rad/s


b) Through what total angle does the spool rotate whileslowing to a stop?
c) What is the angular acceleration during this motion?

I don't really know how to do the rest of the parts.

 

[dinospamoni]

For b) what is the circumference of the spool? How far did it go?

 

[Ampere]

For part b, we are given that the spool travels 3m. The spool rolls along its circumference, right? How many circumferences in 3m? (And how many radians in one circumference)?

 

[Simon Bridge]

a. this is correct - you used v_T=rw ... but do you know why this is the case?

b. Get something round and roll it back and forth across a flat surface.
Watch carefully.

What is the relationship between the distance rolled and the amount the object turns?

c. How would you find regular acceleration?
Relationship between the linear acceleration of the edge of the ball and it's angular acceleration is...

 

[Sneakatone]

the circumference is 0.094 m, and it traveled 3 m

there 95.74 circumferences in 3 m

 

[Ampere]

I got the same circumference, but found a different number of circumferences in 3m. How did you calculate 95.74?

Once you have the # of circumferences, you can use the # of radians in one circumference to find the total angle.

 

[Sneakatone]

3/0.094=95.74
should it have been smaller?

 

[Ampere]

3/0.094 isn't 95.74. Are you making a typo somewhere?

 

[Sneakatone]

I must have had the wrong numbers some how.
my new answer is 31.91

 

[Ampere]

Yes, and how many radians in each circumference?

 

[Sneakatone]

i did 31.91/1.5=21.27 rad.

 

[Ampere]

Aren't there 2pi radians per circumference?

 

[Sneakatone]

31.91/(2pi)=5.07

 

[Ampere]

Other way around. If you have 31.91 circumferences, and 2pi radians/circumference, then you can just multiply.

 

[Sneakatone]

31.91*2pi=200.4

 

[Ampere]

Yes, though if you used unrounded numbers, you'd get exactly 200.

 

[Sneakatone]

so 200 would be the total angle rotated?

 

[Ampere]

In radians, yeah.

 

[Sneakatone]

for part c I know how to solve if there was an extra time given .
would I find time by 3/1.5=2

200/2^2=50 to get the answer?

 

[Sneakatone]

would the equation Ø = w1 t + 1/2 ã t^2 be applied where

Ø - angular displacement
w1 - initial angular velocity
t - time
ã - angular acceleration
??

 

[haruspex]

would the equation Ø = w1 t + 1/2 ã t^2 be applied where

Ø - angular displacement
w1 - initial angular velocity
t - time
ã - angular acceleration
??

If the angular acceleration is constant, yes. (I've not read the OP.)

 

[Sneakatone]

my method on #19 was incorrect ,how should I approach this?

 

[haruspex]

my method on #19 was incorrect ,how should I approach this?

I think the whole question would have been simpler if you'd calculated the time to stop first.
You have an initial and final speed and a distance travelled. Do you know a kinematic equation relating those three to time when the acceleration is constant? If you don't, think what the average speed must be.

 

[Sneakatone]

should the time have been caluculated from the circumference where
1.5m/(1.5*2pi)=0.159 s ?

 

[haruspex]

should the time have been caluculated from the circumference where
1.5m/(1.5*2pi)=0.159 s ?

No, don't worry about the rotation for this part. You know the initial linear speed, the final linear speed and the linear distance. The linear acceleration is constant.

 

[Sneakatone]

the linear initial speed is 1.5 the linear distance is 3, would 1.5 also be the final linear speed?
I feel like 3/1.5=2 should be the time.

 

[haruspex]

the linear initial speed is 1.5 the linear distance is 3, would 1.5 also be the final linear speed?

It comes to rest, right?

 

[Sneakatone]

yes it does come to rest at 3m

 

[haruspex]

yes it does come to rest at 3m

So what's the final linear speed?

 

[Sneakatone]

will it be 1.5 m/s or zero?

 

[haruspex]

will it be 1.5 m/s or zero?

Which of those represents "at rest"?

 

[Sneakatone]

zero is at rest

 

[haruspex]

Right. So there is uniform deceleration from 1.5 m/s to 0 over a distance of 3m. How long does that take?

 

[Sneakatone]

2 seconds

 

[haruspex]

2 seconds

No, it would take 2 seconds to cover 3m at a constant speed of 1.5 m/s, but it it slowing down, so will take longer. What kinematic equation do you know involving two speeds, distance and time when acceleration is constant? If you don't know any such, what kinematic equations do you know?

 

[Sneakatone]

x=vt+1/2at^2

 

[haruspex]

x=vt+1/2at^2

That one involves distance, initial speed, time and acceleration. At this stage, you don't know the acceleration, but you do know the final speed, so you're looking for a different equation. What others do you know?

 

[Sneakatone]

vf^2=vi^2+2add=[(vi+vf)/2]*t

 

[haruspex]

Good, so which of those involves three quantities that you know and the quantity you're trying to calculate?

 

[Sneakatone]

the 1st equation , rearranging it making
a=(vf^2-vi^2)/2d

 

[haruspex]

the 1st equation , rearranging it making
a=(vf^2-vi^2)/2d

We're trying to calculate the time, not the acceleration.

 

[Sneakatone]

so t=d/[(vi+vf)/2]

 

[haruspex]

Right. So using that, how long will it take to come to rest?