I did get the correct answer after subtracting one..I was unsure as to whether the problem-solving was correct of if I got to the answer by chance. Thank you very much for your help!
is that the right way to solve the problem then? Since it asks for how many Earth radii from the surface of the earth..and I included one radius of the eart in my answer..all I do is minus 1 to get the answer?
Thanks.
Homework Statement
The moon has a mass M = 7.36e+022 kg and an average radius R = 1740 km.
for part a)I solved for the gravitational acceleration on the surface of the moon= 1.62 m/s^2
An object on Earth weighs 64 N.
b) What is the weight of this same object on the moon? Answer...
So for part d, it's change in potential energy = just rotational energy then because the system isn't moving translationally?
change in mgh= 1/2 I (omega)^2
h= sin (90-37)* (L/2)= 1.333, and since we know I from part c, we can solve for omega. angular velocity then equals 1.1655 rad/s...
Oh! It should be 37 + 90 degrees.
So you would end up with 133.07 instead of 100.273. Thank you!
Can you help me with part d of the question? I know you need to use conservation of energy.
KE= 0.5I(omega squared)= change in PE(of rod and weight)
Do I use the angle to calculate the...
- Where does the weight of the rod act? What torque does it produce?
The weight of the rod acts at its center of mass so at the pivot point? It produces 0 torque because it's 0m away from the axis of rotation?
- Where does the weight of the mass act? What torque does it produce?
The...
I think I have a hard time deciding what force to use. If it's at an angle, is it right to use the force in the y direction (due to gravity) and add that to the force in the y direction (or y component of force) of the weight and combine them?
For part d,
KE= 0.5I(omega squared)= change...
Homework Statement
A uniform rod is pivoted at its center and a small weight of mass M = 5.07 kg is rigidly attached to one end. You may neglect the size of the weight and assume it is located right at the end of the rod.
a) What vertical force, Fv, must be applied to the opposite end to...