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# Help on angular acceleration problem given mass and angle?

A rod with mass M = 1.3 kg and length L = 0.98 m is mounted on a central pivot. A metal ball of mass m = 0.70 kg is attached to one end of the rod. You may treat the metal ball as a point mass. The system is oriented in the vertical plane and gravity is acting. The rod initially makes an angle θ = 26° with respect to the horizontal. The rod is released from rest. What is the angular acceleration of the rod immediately after it is released?

Picture to go with problem: http://i284.photobucket.com/albums/ll28/bathtub200...

α = _____ rad/s2

### 3 Answers

- SteveLv 71 decade agoFavorite Answer
I = Ir + Ib = (1/12)ML² + m*R² = .272 kg∙m²

α = T/I = mgx/I = mg(.49cos26°)/I = 11.107 rad/s²

- Anonymous4 years ago
The above is incorrect, no longer in basic terms because he did not convert diameter to radius, yet linear acceleration is stress/mass so it may be 100N/4.00kg giving you 25 m/s^2. it truly is an same for section B besides, because the mass and the stress do not replace.

- Anonymous1 decade ago
x = mc2 x 3.14 = a+b+c= the sqare root of nebula