A rod of length 55.0 cm and mass 1.50 kg is suspended by two strings which are 3

Question

A rod of length 55.0 cm and mass 1.50 kg is suspended by two strings which are 37.0 cm long, one at each end of the rod.

1. The string on side B is cut. Find the magnitude of the initial acceleration of end B.

The string on side B is retied and now has only half the length of the string on side A.

*note: The initial linear acceleration is related to the initial angular acceleration, alpha Use the torque about the uncut end, and Newton's 2nd Law for angular motion.

2. Find the magnitude of the initial acceleration of the end B when the string is cut.

*note: the initial linear accel. is related to the initial angular accel., alpha Use the torque about the uncut end, and Newton's 2nd Law for angular motion.

Explanation / Answer

part a )

moment of inertia axis at end = mL^2/3 = 1.5 x 0.55^2/3

I = 0.15125

torque = I*alpha

mg*L/2 = 0.15125 *alpha

alpha = 26.73 rad/s^2

magnitude of initial accelration

a = 0.55 x 26.73 = 14.7 m/s

part b )

angle = cos^-1(0.185/0.55)

angle = 70.3 degree

torque after string is cut

mgsintheta*L/2 = I*alpha

1.5 x 9.8 x sin70.3 * 0.55/2 = 1.5 x 0.55^2/3 * alpha

alpha = 25.16 rad/s^2

magnitude = 25.16 x 0.55

= 13.84 m/s^2

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