A meter stick with a mass of 0.155 kg is pivoted about one end so it can rotate without friction about a horizontal axis. The me
ter stick is held in a horizontal position and released. As it swings through the vertical, calculate:
a. change in gravitational potential energy that has occurred;
b. the angular speed of the stick;
c. the linear speed of the end of the stick opposite the axis.
d. Compare the answer in part (c) to the speed of a particle that has fallen 1.00m, starting from rest.
a. change in gravitational potential energy that has occurred;
b. the angular speed of the stick;
c. the linear speed of the end of the stick opposite the axis.
d. Compare the answer in part (c) to the speed of a particle that has fallen 1.00m, starting from rest.
1 answer:
You might be interested in
Answer:
A) 1.67 x 10 ⁻⁶ m/s
B)5.59 x %
Explanation:
A)
Given:
d = 5.0 km,
mₐ = 2.5 x kg
u₁ = 4.0 x 10⁴ m/s
= 5.98 x 10 ²⁴ kg
Solve using kinetic conserved energy
mₐ x u₁ + x u₂ = uₓ x (mₐ +
)
(2.5 x ) (4.0 x 10⁴ )+ (5.98 x 10 ²⁴ )(0) = uₓ x (2.5 x
+ 5.98 x 10 ²⁴ )
uₓ = ( 2.5 x x 4.0 x 10⁴ ) / (2.5 x
+ 5.98 x 10 ²⁴ )
uₓ = 1.67 x 10 ⁻⁶ m/s
B) Assuming earth radius as a R = 1.5 x 10 ¹¹ m
t = 365 days x 24 hr / 1 day x 60 minute / 1 hr x 60s / 1 minute = 31536000 s
t = 31536000 s
D = 2 π R = 2 π( 1.5 x 10 ¹¹ )
D = 9.4247 x 10 ¹¹ m
u₂ = D / t = 9.4247 x 10 ¹¹ / 31536000
u₂ = 29885.775 m/s
% = ( 1.67 x 10 ⁻⁶ m/s ) / (29885.775 m/s) x 100
% = 5.59 x %