Monday

July 28, 2014

July 28, 2014

Posted by **Elisa** on Monday, July 7, 2008 at 9:34am.

10.0 g object:

15.0g object:

2. A dentist's drill starts from rest. After 3.00 s of constant angular acceleration it turns at a rate of 2.00 multiplied by 104 rev/min.

(a) Find the drill's angular acceleration.

(b) Determine the angle (in radians) through which the drill rotates during this period.

- physics -
**Damon**, Monday, July 7, 2008 at 10:04amwell since everything is in cm and seconds and grams we can stay in the centimeter, gram, second or"cgs" units.

Initial momentum = 10*25 - 15*36

= -290 (to the right is positive)

final momentum =same = -290 = 10 U + 15 V

elastic so initial KE = final KE

(1/2) 10 (625) + (1/2) 15 (1296)

3125+9720

=12,845 dynes

final KE =12,845 = 5 U^2 + 7.5 V^2

so two equations, two unknowns:

-290 = 10 U + 15 V

12,845 = 5 U^2 + 7.5 V^2

U = - (1.5 V + 29)

U^2 = 2.25 V^2 + 87 V + 841

12,845 = 5(2.25 V^2 + 87 V + 841) + 7.5 V^2

You can solve the quadratic I think

- physics -
**Damon**, Monday, July 7, 2008 at 10:10amin 3 seconds from 0 to 2*10^4 revs/min

first change rev/min to radians/sec

2*10^4rev/min (1 min/60 s)(2 pi rad/rev)

=2094 radians/s

alpha = change in speed/change in time = 2094/3 = 698 radians/s^2

angle = (1/2) alpha t^2

=349*9 = 3141 radians

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