Question

a. A rotating wheel requires 7.00 s to rotate 25.0 revolutions. Its angular velocity at the end of the 7.00 s interval is 97.0 rad/s. What is the constant angular acceleration of the wheel?

b. A machine part rotates at an angular speed of 0.52 rad/s; its speed is then increased to 3.2 rad/s at an angular acceleration of 0.52 rad/s². Find the angle through which the part rotates before reaching this final speed.

c. A dentist's drill starts from rest. After 3.40 s of constant angular acceleration, it turns at a rate of 2.70 ✕ 10⁴ rev/min.

(a) Find the drill's angular acceleration- rad/s²
(b) Determine the angle (in radians) through which the drill rotates during this period.

Answer 1

t=75 = 25 rev --Son rad CI glov = 271 wad) w = 97 dod/s Let initial angular velocity angular acceleration be a be w. 4 wwo txt 973w6+72 0 O= wot + ext? Son=7W, +49X → and then Multiplying co by 7 subtekefing Ø 4☺ 7wo + 494 = 679 -747-49% = -50% - 49x = 521.9.2. x= 21.30 scad/s? WN 0.52 oladis w = 3.2 oradys x=0-52 cradis? - W2W2 = 2x0 8= w2_wool 22 Scanned with - 9.59 orad Scanner

c) a - w =0 t = 3.4.5. W = 2:7x10 reunin 3227x10x20 - Orad/s W = 2.83X10 aad/snood new-wong l ittle 219.3x10'co felii L rotan 3.4 istent = 8.32X10crad/s? 2 il So swt +fXt? 0 =0+448.3 2X10*x( 3-4) fo Hittaja O=4.81 x103 read i t cc Scanned with CamScanner