The correct answers are shown but I\'d like to know how to get to that answer, s

Question

The correct answers are shown but I'd like to know how to get to that answer, someone please help!

A rod of length 53.0 cm and mass 1.40 kg is suspended by two strings which are 42.0 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. 1.47x10^1 m/s^2 (in m/s^2) 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. 1.35x10^1 m/s^2 (in m/s^2)

Explanation / Answer

initially the rod is in equillibrium

BY symmetry, let the tenstion be T

net force 2T = mg

after the string ifs cut, the rod rod rotates due to torque of mg

so

Torque of mg about A = mgL/2

acclerationa t end B = T/Inerti

= mgl/2 /, ml^2/3

= 3g/2L

so

a = 3*9.8/(2)

a = 1.47 *10^1   option G it is

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2:

let x ne the angle ofm inclination before the string is cut

so

sin theta   = halff of the length of the string./length of the rod

sin xtheta = 24/53 = 0.452

x = 27 deg

now when the string is cut,

mg acts downwards and T acts upwards

so

rod rotaes about point A to torque of mg

so

Ang accleration a = T/I

a a= mgl cos theta/2   * 3/ml^3

= mg cos theta /2L

hence accleration a = Ang acc * diusatnce

= mg cos theta/2 /mL^2/3

= 3g cos theta /2L * L

= 3* 9.81 * cos 27/2

= 1.34 *10^1 m/s^2 ---->>>>>>>>>>>>>>>>option G it is

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