A meterstick (L = 1 m) has a mass of m : 0.231 kg. Initially it hangs from two s

Question

A meterstick (L = 1 m) has a mass of m : 0.231 kg. Initially it hangs from two short strings one at the 25 cm mark and one at the 75 cm mark. 1) What is the tension in the left string? NSubmit You currently have 0 submissions for this question. Only 10 submission are allowed You can make 10 more submissions for this question. 2) Now the right string is cut! What is the initial angular acceleration of the meterstick about its pivot point? (You may assume the rod pivots about the left string, and the string remains vertical) rad/ Submit You currently hove O submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question. 3) What is the tension in the left string right after the right string is cut? N Submit You currently have O submissions for this question. Only 10 submission are allowed. You can moke 10 more submissions for this question. 4) After the right string is cut, the meterstick swings down to where it is vertical for an instant before it swings back up in the other direction. What is the angular speed when the meterstick is vertical? rad/s Submiat You currently have 0 submissions for this question. Only 10 submission are allowed.

Explanation / Answer

1] Because of symmetry the tension in each string will be equal,

2T = mg

T = 0.5 mg = 0.5*0.231*9.8 = 1.132 N

2] angular acceleration = torque/moment of inertia

= (mg*x)/(1/12 mL^2 + mx^2)

= 9.8*0.25/(1/12*1^2 + 0.25^2)

= 16.8 rad/s^2

3] acceleration a = alpha *x = 16.8*0.25 = 4.2 m/s^2

mg - T = ma

T = mg-ma = 0.231*(9.8-4.2)

= 1.2936 N answer

4] by energy conservation, 0.5iw^2 = mgh

w = sqrt(2mgh/i) = sqrt(2mgh/(1/12 mL^2 + mx^2))

= sqrt(2*9.8*0.25/(1/12*1^2 + 0.25^2))

= 5.8 rad/s

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