A dentist's drill starts from rest. After 3.20 s of constant angular acceleration, it turns at a rate of 2.51x103 rev/min. (a) Find the drill's angular acceleration. (b).Determine the angle (in radians) through which the drill rotates during this period.

Starting from rest at t = 0, a wheel undergoes a constant acceleration from t = 0 to t = 10 s. When t = 3.0 s, the angular velocity of the wheel is 6.0 rad/s. Through what angle does the wheel rotate from t = 0 to t = 20 s.

The angular position of a point on a rotating wheel is given by 0 = 7.67+3.84t2 +3.65t3, where e is in radians and t is in seconds. At t= 0, what are (a) the point's angular position and (b) its angular velocity? (c) What is its angular velocity at t 8.10 s? (d) Calculate its angular acceleration at t = 2.83 s. (e) Is its angular acceleration constant? (a) Number Units (b) Number Units (c) Number i Units (d) Number i Units (e) Number

An electric drill starts from rest and undergoes uniform angular acceleration for a period of 0.130 s until it turns at a rate of 2.35 x 10 rev/min. (a) What is the drill's angular acceleration (in rad/s)? (Enter the magnitude.) rad/s? (b) What is the angle (in radians) through which the drill rotates during this period? (Enter the magnitude.) rad

A disk, initially rotating at 106 rad/s, is slowed down with a constant angular acceleration of magnitude 2.35 rad/s2. (a) How much time does the disk take to stop? (b) Through what angle (rad) does the disk rotate during that time?

A wheel rotates with a constant angular acceleration of 3.50 rad/s2.(A) If the angular speed of the wheel is 2.00 rad/s at ti = 0, through what angular displacement does the wheel rotate in 2.00 s?                                      (B) Through how many revolutions has the wheel turned during this time interval?                                                                                                              (C) What is the angular speed of the wheel at t = 2.00 s?

A machine part rotates at an angular speed of 0.066 rad/s; its speed is then increased to 2.3 rad/s at an angular acceleration of 0.80 rad/s2. (a) Find the angle through which the part rotates before reaching this final speed. rad (b) If both the initial and final angular speeds are doubled and the angular acceleration remains the same, by what factor is the angular displacement changed? by a factor of 2 by a factor of 4 by a factor of 16 The angular displacement doesn't change.

A dentist causes the bit of a high-speed drill to accelerate from an angular speed of 1.79 x 10^4 rad/s to an angular speed of 3.13 × 10^4 rad/s. In the process, the bit turns through 3.01 x 10^4 rad. Assuming a constant angular acceleration, how long would it take the bit to reach its maximum speed of 8.46 × 10^4 rad/s, starting from rest?

A person is riding a bicycle, and its wheels have an angular velocity of 16.1 rad/s. Then, the brakes are applied and the bike is brought to a uniform stop. During braking, the angular displacement of each wheel is 12.7 revolutions. (a) How much time does it take for the bike to come to rest? (b) What is the anguar acceleration (in rad/s2) of each wheel?

The rotor of a helicopter is gaining angular speed with constant angular acceleration. At t = 0 it is rotating at 1.25 rad/s. From t = 0 to t = 2.00 s, the rotor rotates through 8.00 rad. (a) What is the angular acceleration of the rotor? (b) Through what angle (in radians) does the rotor rotate from t = 0 to t = 4.00 s?