Rotation angular velocity of a turntable

In summary, the conversation is about finding the turntable's rotational angular velocity when a block reaches its outer edge. The process involves using the conservation of momentum equation and the equation for moment of inertia for the turntable and block. The initial attempt using conservation of energy resulted in an answer of 5.52 rad/s, which was close but not quite the correct answer. After rechecking the equation for the moment of inertia of the turntable, a final answer of 5.42 rad/s was obtained. The conversation also suggests double-checking calculations to minimize errors.
  • #1
lilyE
5
0
I've spent at least an hour trying to figure this out, but can't seem to figure out how to solve this.

1. Homework Statement

A 170g , 30.0-cm-diameter turntable rotates on frictionless bearings at 66.0rpm . A 25.0g block sits at the center of the turntable. A compressed spring shoots the block radically outward along a frictionless groove in the surface of the turntable.

What is the turntable's rotation angular velocity when the block reaches the outer edge?

Homework Equations


Conservation of momentum: Ii*wi = If*wf

The Attempt at a Solution


I(turntable) = Mr^2 = (.170 kg)(.15 m)^2 = 3.825*10^-3 kg m^2
I(block)o = mr^2 = 0
I(block)f = mr^2 = (.025 kg)(.15 m)^2 = 5.265*10^-4 kg m^2
wi = ((66 rpm)(2pi))/60 = 6.91 rad/s
I(turntable+block)f = 3.825*10^-3 + 5.265*10^-4 = 4.3875*10^-3 kg m^2

(3.825*10^-3)(6.91) = (4.3875*10^-3)wf
0.1826 = (4.3875*10^-3)wf
wf = 6.02 rad/s -- wrong

Initially, I tried using conservation of energy, and ended up with 5.52 rad/s. The program told me "not quite" and suggested a rounding error, so I think 5.52 might be close to the right answer, but I can't seem to figure out why I'm not getting it. Any help would be much appreciated!
 
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  • #2
I think you should recheck the equation of ##I_{turntable}##.
 
  • #3
Would it just be 1/2*(.170)(.15)? That's what I had at first, but I wasn't sure if that was correct or not.
 
  • #4
If I use Iturntable = (0.5)(0.17)(0.15)2, I get Iturntable = 1.9125 * 10-3 kg m2. Plugging this in, I get a final answer of 5.42 rad/s. Does this seem right?
 
  • #5
Believe yourself.
 
  • #6
By the way, due to the error generated by calculating, recommend you to calculate at the last pace.
You can get the best result.
 

Related to Rotation angular velocity of a turntable

What is the definition of rotation angular velocity?

Rotation angular velocity is defined as the rate of change of angular displacement with respect to time. It measures how quickly an object is rotating and is typically measured in radians per second (rad/s).

How is rotation angular velocity different from linear velocity?

Rotation angular velocity measures how quickly an object is rotating around a fixed axis, while linear velocity measures how quickly an object is moving in a straight line. Rotation angular velocity is measured in radians per second, while linear velocity is measured in meters per second.

What factors affect the rotation angular velocity of a turntable?

The rotation angular velocity of a turntable can be affected by the mass and distribution of weight on the turntable, the power and torque of the motor, and any external forces or friction acting on the turntable.

How is rotation angular velocity calculated?

Rotation angular velocity can be calculated by dividing the change in angular displacement by the change in time. It can also be calculated by dividing the linear velocity at a point on the turntable by the distance from the center of rotation.

Why is rotation angular velocity important in turntables?

Rotation angular velocity is important in turntables because it determines the speed at which the record is rotated, which directly affects the pitch and sound quality of the music being played. It is also important in maintaining the stability and balance of the turntable while in use.

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