A rod of length 50.0 cm and mass 1.20 kg is suspended by two strings which are 4

Question

A rod of length 50.0 cm and mass 1.20 kg is suspended by two strings which are 40.0 cm long, one at each end of the rod.

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

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

A rod of length 50.0 cm and mass 1.20 kg is suspended by two strings which are 40.0 cm long, one at each end of the rod. 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. Find the magnitude of the initial acceleration of the end B when the string is cut.

Explanation / Answer

Torque after string is cut
= 1.20 g * 0.50 / 2
= 2.943 Nm
= M of I * angular accleration
= 1.20 * 0.50^2 / 3 * A
= 0.1 A

0.1 A  = 2.943
A = 29.43 rad /s^2

the magnitude of the initial acceleration of end B
= 29.43 * 0.50
= 14.71 m/s^2

A.
angle made with vertical
= arc cos 0.2 / 0.50
= 66.42 deg

Torque after string is cut
= 1.20 g * 0.50 sin (66.42 / 2)
= 3.223 Nm
= M of I * angular accleration
= 1.20 * 0.50^2 / 3 * A
= 0.1 A
0.1 A = 3.223
A = 32.23 rad /s^2
magnitude of the initial acceleration of the end B when the string is again cut

=32.23 * 0.50
= 16.115 m/s^2

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