Calculate Centripetal Force for Object on Turntable

[Flinthill84]

a small object is placed 10cm from the center of a phonograph turntable. it is observed to remain on the table when it rotates at 33 1/3 rev/min but slides off when it rotates at 45 rev/min. Calculate the centripetal force.


if anyone has any ideas what so ever they are greatly appreciated as i am all out.

 

[UrbanXrisis]

use: Fc=4π2mr/T
however, you need a mass. Force is mass times acceleration. Without a mass, I'm not sure how you would find a force.

 

[wais]

To calculate the centripetal force, we can use the formula Fc = mv^2/r, where Fc is the centripetal force, m is the mass of the object, v is the velocity, and r is the radius of rotation.

First, we need to convert the given rotation rates into angular velocity. 33 1/3 rev/min can be converted to 33.33 rev/min by dividing by 60 seconds. This is equal to 0.5555 rev/s. Similarly, 45 rev/min can be converted to 45/60 = 0.75 rev/s.

Next, we need to calculate the velocity of the object at each rotation rate. Since the object is placed 10cm from the center, the radius of rotation (r) is 10cm or 0.1m. At 33.33 rev/s, the velocity (v) can be calculated as v = 2πr(0.5555) = 3.48 m/s. At 0.75 rev/s, the velocity is v = 2πr(0.75) = 4.71 m/s.

Now, we can plug these values into the formula to calculate the centripetal force. At 33.33 rev/s, Fc = (m)(3.48)^2/(0.1) = 12.11m N. At 0.75 rev/s, Fc = (m)(4.71)^2/(0.1) = 22.19m N.

Therefore, the centripetal force required to keep the object on the turntable at 33.33 rev/s is 12.11m N, and at 0.75 rev/s is 22.19m N. This shows that as the rotation rate increases, the centripetal force needed to keep the object on the turntable also increases. This is because the velocity of the object increases, and according to the formula, the centripetal force is directly proportional to the square of the velocity.