Calc Centripetal Acceleration: Rotational Motion & Vinyl Record Player

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nbroyle1
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In the days before compact discs and MP3 players (ancient history!), music was recorded in scratches in the surface of vinyl-coated disks called records. In a typical record player, the record rotated with a period of 3.6 s. Find the centripetal acceleration of a point on the edge of the record. Assume a radius of 8.89 cm.

So I already converted the radius to meters and got .0889m.

Now I need to find the velocity in order to plug it into the equation for centripetal acceleration which is Ac=v^2/r...

How should I go about solving for the velocity so I can get the centripetal acceleration??
 
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nbroyle1 said:
In the days before compact discs and MP3 players (ancient history!), music was recorded in scratches in the surface of vinyl-coated disks called records. In a typical record player, the record rotated with a period of 3.6 s. Find the centripetal acceleration of a point on the edge of the record. Assume a radius of 8.89 cm.

So I already converted the radius to meters and got .0889m.

Now I need to find the velocity in order to plug it into the equation for centripetal acceleration which is Ac=v^2/r...

How should I go about solving for the velocity so I can get the centripetal acceleration??
What distance does that point on the edge travel in one second?
 
im not sure... So its traveling 2∏ in 3.6s right??
 
360 degrees. So then in one second I think it travels 100 degrees??
 
So I found that the acceleration is 1.745rad/s^2. Then I plugged it into the equation for average acceleration to find the velocity and got 6.28m/s... is this correct??
 
nbroyle1 said:
So I found that the acceleration is 1.745rad/s^2. Then I plugged it into the equation for average acceleration to find the velocity and got 6.28m/s... is this correct??
I don't think so.

What is the circumference of the circle made by the point on the edge?
 
.1778m is the circumfrance
 
how does the circumference help me I am confused??