Angular acceleration of a drill

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The discussion revolves around calculating the angular acceleration of a dentist's drill that starts from rest and reaches a speed of 2.20 x 10^4 rev/min after 2.90 seconds. The user initially converts the rotational speed to rad/sec but encounters an error when calculating angular acceleration, arriving at an incorrect value of 886.09 rad/sec^2. A suggestion is made to double-check the time used in the calculations, emphasizing the importance of accuracy in division. The user expresses frustration over the mistake but appreciates the guidance. The conversation highlights the common challenges faced in physics problem-solving.
Dr_bug
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Homework Statement


A dentist's drill starts from rest. After 2.90 s of constant angular acceleration it turns at a rate of 2.20 x 10^4 rev/min.
(a) Find the drill's angular acceleration (in rad/sec^2).
(b) Determine the angle (in radians) through which the drill rotates during this period.

Homework Equations





The Attempt at a Solution


okay so for (a) I thought that I should convert the rev/ min to rad/sec first and got 2303.8 rad/sec and then I used: change in angular velocity/ change in time and got 886.09 but when I input this answer, webassign (thats how we submit homework) says I'm wrong... I don't know what else to try
 
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Dr_bug said:

Homework Statement


A dentist's drill starts from rest. After 2.90 s of constant angular acceleration it turns at a rate of 2.20 x 10^4 rev/min.
(a) Find the drill's angular acceleration (in rad/sec^2).
(b) Determine the angle (in radians) through which the drill rotates during this period.

Homework Equations





The Attempt at a Solution


okay so for (a) I thought that I should convert the rev/ min to rad/sec first and got 2303.8 rad/sec and then I used: change in angular velocity/ change in time and got 886.09 but when I input this answer, webassign (thats how we submit homework) says I'm wrong... I don't know what else to try

Are you using 2.9 s for your time? Check your division.
 
Ahh Geez... thanks I was going crazy
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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