Questions 41-42
B. 800 J
C. 4000 J
D. 9600 J
E. 19,200 J
Solution:
Rotational Kinetic Energy:
&space;=&space;\frac{1}{2}\cdot&space;4\left&space;(40^2-80^2&space;\right&space;)&space;=&space;-9600&space;\text{&space;J} "\Delta T_\text{rot} = \frac{1}{2}I\left (\omega_f^2-\omega_i^2 \right ) = \frac{1}{2}\cdot 4\left (40^2-80^2 \right ) = -9600 \text{ J}")
Answer: D
A cylinder with moment inertia 4 kgm² about a fixed axis initially rotates at 80 radians per second about this axis. A constant torque is applied to slow it down to 40 radians per second. The kinetic energy lost by the cylinder is
A. 80 JB. 800 J
C. 4000 J
D. 9600 J
E. 19,200 J
(GR9277 #41)
Solution:
Rotational Kinetic Energy:
Answer: D
No comments :
Post a Comment