String speed and tension

  • Thread starter Kchu
  • Start date
  • #1
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A ball of mass m is attached to a string of length . It is being swung in a vertical circle with enough speed so that the string remains taut throughout the ball's motion. Assume that the ball travels freely in this vertical circle with negligible loss of total mechanical energy. To avoid confusion, take the upward direction to be positive throughout the problem. At the top and bottom of the vertical circle, label the ball's speeds V_t and V_b, and label the corresponding tensions in the string T_t and T_b

Find T_b-T_T, the difference between the tension in the string at the bottom relative to that at the top of the circle.

Express the difference in tension in terms of m and g

The tension in the bottom i got was

T_b=v_b^2/L*m+(m*g)

the tension in the top i got was

T_t=m(-v_t^2/L)-m*g

to relate these and the total mech i did
v_t^2 =v_b^2-4*g*L

and i solved and i got 4*m*g and it marked it wrong saying "Check the energy difference between top and bottom in your calculation." can anyone help?
 

Answers and Replies

  • #2
mukundpa
Homework Helper
524
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1.
T_t=m(-v_t^2/L)-m*g
This should be written as
T_t=m[(-v_t)^2/L]-m*g
2.
T_b - T_t = [v_b^2/L*m+(m*g)] - [m(-v_t^2/L)-m*g]=(m/L)(v_b^2 - V_t^2) +2mg
 
  • #3
10
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Your equations were right but you solved them incorrectly, i think the correct answer is 6mg.
 
  • #4
mukundpa
Homework Helper
524
3
o man ! I have not solved the equations but indicated the mistakes and written the equation as the value of (v_b^2 - V_t^2) is already calculated by Kchu as 4gL, sbstituting this in equation you will get your answer.
 

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