A dentist\'s drill starts from rest. After 2.70 s of constant angular accelerati

Question

A dentist's drill starts from rest. After 2.70 s of constant angular acceleration it turns at a rate of 2.45 times 10^4 rev/min. (a) Find the drill's angular acceleration. (b) Determine the angle (in radians) through which the drill rotates during this period. A bicycle is turned upside down while its owner repairs a flat tire. A friend spins the other wheel and observes that drops of water fly off (see figure below). A drop that breaks loose from the tire on one turn rises vertically 59.0 cm above the tangent point. A drop that breaks of the wheel is 0.315 m.

Explanation / Answer

w=w0 + at

24500/60=0+a*2.7

a=151.23 rev/s2

w2 =w02 + 2a(theta)

(24500/60)2= 0 +2*151.23*theta

thetaa=551.266 rev=551.266/2pi rad=87.74 ras

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