Kinematics of Rotational Motion Part II

In summary: No, gear B/Drum C are not at rest and are not accelerating uniformly. Gear A is accelerating uniformly from rest making 20 revolutions in 12secs.
  • #1
freshbox
290
0

Homework Statement


I cannot get the answer for part c, -13.02rad/s^2


The Attempt at a Solution



52.08=260.42+α(10)
-208.34=10α
α= -20.83rad/s^2 (Correct)


0=52.08+α(10)
-52.08=10α
α=-5.208rad/s^2 (Wrong)


Please help, thanks..
 

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  • #2
How long did it take to come to rest? (How long was it moving at constant speed?)
 
  • #3
hifreshbox! :smile:

(try using the X2 button just above the Reply box :wink:)
freshbox said:
52.08=260.42+α(10)
-208.34=10α
α= -20.83rad/s2(Correct)


0=52.08+α(10)
-52.08=10α
α=-5.208rad/s2 (Wrong)

the 10 seconds for 300 to 60 mph is correct

the 10 seconds for 60 to 0 mph is not correct …

you need to allow for the time spent cruising :wink:
 
  • #4
How long did it take to come to rest? (How long was it moving at constant speed?)

Since it states that total is 20sec, 300km/h-60km/h is 10sec, so 60km/h-0km/h is 10sec since 20-10=10
 
  • #5
but in between, "it travels for another 100 m at the constant speed of 60 km/h" !
 
  • #6
but it says
"and it takes a total of 20 seconds to stop at the pit from its top speed of 300km/h"

total =20 secs?
 
  • #7
freshbox said:
but it says
"and it takes a total of 20 seconds to stop at the pit from its top speed of 300km/h"

total =20 secs?
Yes, the total time is 20 seconds. But that includes three segments:
- initial slowdown to intermediate speed
- constant speed motion at intermediate speed
- final slowdown to zero speed

Do not ignore the 'constant speed' segment. How much time is spent there?
 
  • #8
Is 2nd segment vi=16.67m/s v=16.67m/s a=0m/s^2 ?
 
  • #9
How do i calculate the time when my acceleration is 0 for segment b?

ω=ωi+αt,v=vi+at
a times t = 0

No t in the equation anymore..:redface::cry:
 
  • #10
It's constant velocity motion: distance = speed * time.
 
  • #11
my book only has this formula.. no wonder i keep getting stucked..

So S=d/t is the only way to solve?

Thanks btw Doc Al
 

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  • #12
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  • #13
freshbox said:
my book only has this formula.. no wonder i keep getting stucked..
What formula? The attachment was blank.
So S=d/t is the only way to solve?
That's the basic definition of speed, so sure.
 
  • #14
The equation you want from your list is:
s = vit + 1/2at2

(s is distance, of course.)

Just plug in a = 0.
 
  • #15
Ah... yea why didn't i see that hehe.. i understand already, thanks for the help..

Btw if you don't mind can you take a look at https://www.physicsforums.com/showthread.php?t=605830..I just want to clarify some doubts..

Since the question says that "Gear A is accelerated uniformly from rest making 20 revolutions in 12secs"

Gear A: Vi=0, t=12s

Can I say that Gear B/Drum C is also Vi=0, t=12s since they are all connected together.
 

Related to Kinematics of Rotational Motion Part II

What is rotational motion?

Rotational motion is the movement of an object around a fixed point or axis.

What are the key components of rotational motion?

The key components of rotational motion include angular displacement, angular velocity, and angular acceleration.

How is rotational motion different from linear motion?

Rotational motion involves movement around a fixed point, whereas linear motion involves movement in a straight line.

What is the relationship between angular and linear velocity?

Angular velocity is the rate of change of angular displacement, while linear velocity is the rate of change of linear displacement. The two are related by the equation v = rω, where v is linear velocity, r is the radius, and ω is angular velocity.

How is torque related to rotational motion?

Torque is the measure of the force that causes an object to rotate around an axis. It is directly proportional to the angular acceleration of an object and is represented by the equation τ = Iα, where τ is torque, I is moment of inertia, and α is angular acceleration.

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