A particle executes simple harmonic motion with an amplitude of 1.69 cm. At what positive displacement from the midpoint of its
1 answer:
You might be interested in
Answer:
- The gravity does a work of - 117.6 Joules.
- The tension does not do work as the force is perpendicular to the direction of motion at any point in the trajectory.
Explanation:
The work done by the gravity simply is the difference in gravitational potential energy multiplied by -1:
where m is the mass of the ball, g is the acceleration due to gravity, is the final height and
is the initial height.
So, if the radius is 2.00 m, then the difference of height will be 4 meters:
As the tension is perpendicular to the velocity of the ball, the force is always perpendicular to the direction of motion. So, the differential of work will be:
Answer:
w = 1.976 rpm
Explanation:
For simulate the gravity we will use the centripetal aceleration , so:
where w is the angular aceleration and r the radius.
We know by the question that:
r = 60.5m
= 2.6m/s2
So, Replacing the data, and solving for w, we get:
W = 0.207 rad/s
Finally we change the angular velocity from rad/s to rpm as:
W = 0.207 rad/s = 0.207*60/(2)= 1.976 rpm