Rotational kinematics of a turntable

In summary, the electric turntable 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. After 0.192 seconds, the blade has turned 0.57 revolutions. The resultant acceleration at the blade at this time is 1.98 rad/s.
  • #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.
 
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  • #2
Crusaderking1 said:
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.
 
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  • #3
Crusaderking1 said:

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.

Crusaderking1 said:
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
Crusaderking1 said:
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

Crusaderking1 said:
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?
 
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1. What is rotational kinematics?

Rotational kinematics is a branch of physics that focuses on the motion of objects in circular motion, such as a turntable. It involves the study of angular displacement, velocity, and acceleration.

2. How does a turntable rotate?

A turntable rotates due to the force of friction between the stylus and the record. As the stylus moves along the grooves of the record, it creates a torque that causes the turntable to rotate.

3. What is angular velocity?

Angular velocity is the rate at which an object rotates around an axis. It is measured in radians per second and is calculated by dividing the change in angular displacement by the change in time.

4. How does the radius of the turntable affect its rotational kinematics?

The radius of the turntable affects its rotational kinematics by determining its moment of inertia. A larger radius means a larger moment of inertia, which requires more torque to produce a given angular acceleration.

5. What factors can affect the rotational kinematics of a turntable?

The rotational kinematics of a turntable can be affected by factors such as the mass and shape of the turntable, the mass and shape of the record, the friction between the stylus and the record, and any external forces acting on the turntable.

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