A uniform rod of mass M = 1.2 kg and length L = 0.80 m,lying on a frictionless horizontal plane,is free to pivot about a vertical axis through one end,as shown.The moment of inertia of the rod about this axis is given by (1/3) ML2.If a force (F = 5.0 N,θ = 40°) acts as shown,what is the resulting angular acceleration about the pivot point?
A) 16 rad/s2
B) 12 rad/s2
C) 14 rad/s2
D) 10 rad/s2
E) 33 rad/s2
Correct Answer:
Verified
Q23: Particles (mass of each = 0.20 kg)
Q25: A uniform rod (length = 2.0 m)
Q26: A uniform rod is 3.0 m long.
Q28: Identical particles are placed at the 50-cm
Q29: A uniform meter stick is pivoted to
Q32: The rigid object shown is rotated about
Q33: A uniform rod is 2.0 m long.
Q33: If M = 0.50 kg,L = 1.2
Q35: A uniform rod (mass = 1.5 kg)
Q36: A uniform rod is 2.0 m long.
Unlock this Answer For Free Now!
View this answer and more for free by performing one of the following actions
Scan the QR code to install the App and get 2 free unlocks
Unlock quizzes for free by uploading documents