Question

7. A baseball bat starts from rest and is swing through 140 ° in 0.7 s. What is the angular acceleration?

A. 200 °/s2 B. 189.1 °/s2 C. 564.3 °/s2 D. 571.4 °/s2

8. A diver starts from rest and accelerates through a somersault at 950 °/s2. What is the angular displacement over 0.15 s?

A. 10.69° B. 142.5° C. 71.25 °D. 950°

9. A golf club started from rest and reached a final angular velocity of 370 °/s with a constant angular acceleration of 580°/s2. How many degrees did the club rotate?

A. 210 B.118 C156 D.213

Answer 1

Question 7

Initial angular speed of the baseball bat = = 0 ^o/s (At rest)

Time period = T = 0.7 sec

Angle through which the baseball bat swings = = 140 ^o

Angular acceleration of the baseball bat =

= T + T²/2

140 = (0)(0.7) + (0.7)²/2

= 571.4 ^o/s²

Angular acceleration of the baseball bat = 571.4 ^o/s²

Question 8

Initial angular speed of the diver = = 0 ^o/s (At rest)

Angular acceleration of the diver = = 950 ^o/s²

Time period = T = 0.15 sec

Angular displacement of the diver =

= T + T²/2

= (0)(0.15) + (950)(0.15)²/2

= 10.69 ^o

Angular displacement of the diver over 0.15 sec = 10.69 ^o

Question 9

Initial angular velocity of the golf club = ₁ = 0 ^o/s (At rest)

Final angular velocity of the golf club = ₂ = 370 ^o/s

Angular acceleration of the golf club = = 580 ^o/s²

Angle through which the golf club rotated =

₂² = ₁² + 2

(370)² = (0)² + 2(580)

= 118 ^o

Angle through which the club rotates = 118 ^o