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Rotational kinematics of a turntable

  • #1

Homework Statement



An electric turntable 0.710 m in diameter is rotating about a fixed axis with an initial angular velocity of 0.230 rev/s. The angular acceleration is 0.887 rev/s^2.

A. Compute the angular velocity after a time of 0.192 s.

B. Through how many revolutions has the blade turned in this time interval?

C. What is the tangential speed of a point on the tip of the blade at time t = 0.192 s?

D. What is the magnitude of the resultant acceleration of a point on the tip of the blade at time t = 0.192 s?


Homework Equations



Any rotational kinematic.
Conversions to radians/revolutions




The Attempt at a Solution



A.) I took the angular velocity 0.230 and multiplied it by 2pi which equals 1.445 rad/s.

I then plugged it into a kinematic equation, 1.445+(0.887*0.230)^2 = 2.56 rad/s = .400 rad/s.

This is right.

B.) The I used the distance rotational kinematic equation, 1.445 * 0.230 +.5(0.887)(0.230)^2 = 0.3558 rad. = 0.57 rev.

This is wrong. Need help.

C) To to find the tangential speed, I did 0.710 m * 2.56 rad/s = 1.82.

This is wrong. Need help.

D) Need help.

Thanks.
 

Answers and Replies

  • #2
dynamicsolo
Homework Helper
1,648
4
An electric turntable 0.710 m in diameter is rotating about a fixed axis with an initial angular velocity of 0.230 rev/s. The angular acceleration is 0.887 rev/s^2.

A. Compute the angular velocity after a time of 0.192 s.

B. Through how many revolutions has the blade turned in this time interval?


B.) The I used the distance rotational kinematic equation, 1.445 * 0.230 +.5(0.887)(0.230)^2 = 0.3558 rad. = 0.57 rev.

This is wrong. Need help.
On part (b), have you perhaps "plugged in" the initial angular velocity where the time interval 0.192 seconds belongs?

I think you have misled yourself because you have omitted a factor (and wrote another incorrectly) in the typing of your equation in part (A):


I then plugged it into a kinematic equation, 1.445+(0.887*0.230)^2 = 2.56 rad/s = .400 rad/s.
I believe that should be 1.445 + [(0.887*2*pi)^2] (0.192) = 2.56 rad/s . I agree with your result, but you didn't get it using what you've typed.
 
Last edited:
  • #3
rock.freak667
Homework Helper
6,230
31

The Attempt at a Solution



A.) I took the angular velocity 0.230 and multiplied it by 2pi which equals 1.445 rad/s.
Conversion is correct.

I then plugged it into a kinematic equation, 1.445+(0.887*0.230)^2 = 2.56 rad/s = .400 rad/s.
From what I gather, your formula is Ω = ω + (αω)2 ?

I am not following this line as that computation gives me 1.48 or so.
 
  • #4
The angular velocity w = w0 + a*t where w0=0.23*2*Pi rad./s
also where 'a' = 0.887*2*Pi rad./s^2
At t=0.192, w(0.192) = 0.23*2*Pi + 0.887*2*Pi*0.192
=2.5 rad/s

On part B, I need to plug in time(0.192 s) where I plugged in initial angular velocity?

For, C, should i be using the radius of .355 m? Then multiply it by 2.56?
 
  • #5
rock.freak667
Homework Helper
6,230
31
On part B, I need to plug in time(0.192 s) where I plugged in initial angular velocity?
The formula is θ = ωt + ½αt2

You plugged in t as 1.445 and ω as 0.230, you need to plug in ω as 1.445 and t=0.192. Just makes sure α is in rad/s2

For, C, should i be using the radius of .355 m? Then multiply it by 2.56?
Yes that would be correct since you know that v = rω
 
  • #6
Ok, thank you guys for all the help! much appreciated.
 
  • #7
For part D.

I received the acceleration of the radius by multiplying 0.887 rad/s^2 by 2pi.
=5.73 rad/s

I received the tangential acceleration by the change in speed of (2.56 rad/s)^2 multiplied by the radius of 0.355 m.
= 2.195 rad/s.

(5.57)^2+(2.19)^2 and the square root = 5.99 rad/s which is wrong.

The I tried (2.515)^2(.355) = 2.25 = radial acceleration
with (5.573)(0.355) = 1.98

The use Pythagorean theorem for 2.99 m/s^2

Is this right?
 
Last edited:

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