1. A rod of mass M and length L are pivoted so that it can freely rotate about i

Question

1. A rod of mass M and length L are pivoted so that it can freely rotate about its center (I = 1/12 ML^2) a constant force F is applied at one end of the rod, the force vector makes an angle theta with the direction parallel to the rod what is the angular acceleration of the rod?

My Answer: I = 1/12 ML^2, torque = (I) (angular acceleration), torque = (L)(F)sin(theta)
angular acceleration = (torque)/(I) so angular acceleration = ((L)(F)sin(theta)) / (1/12ML^2) is this right?

2. A particle of mass 2.00 kg is 5.00 m away from the origin, its position vector makes a 30.0 degree angle with the positive x axis what is its moment of inertia about the x-axis?

my Answer: I = Mr^2, I = (2.00 kg)(5.00m)^2 = 50 kg *m^2 is this right?

Explanation / Answer

yes..u did right

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