The 5-kg mass at the end of the 4.0 m cord has a tension in the cord of 200 N at

Question

The 5-kg mass at the end of the 4.0 m cord has a tension in the cord of 200 N at the bottom of the arc, ?=0.  Using energy techniques, determine at what angle ? the cord goes slack?  

Hints:  The slack angle is greater than 90

The 5-kg mass at the end of the 4.0 m cord has a tension in the cord of 200 N at the bottom of the arc, ?=0. Using energy techniques, determine at what angle ? the cord goes slack? Hints: The slack angle is greater than 90 degree and less than 120 degree . Draw the generic n-t FBD in approximately the position shown. It can then be used in any angle ?. The velocity at the slack angle is not zero. Use n-t to find an expression for velocity to be used in the energy equation.

Explanation / Answer

there will be centripetal force acting on the mass

Tension will be towards the centre

so

T- 0.5mw^2 *r = mg

200 - 0.5*5*w^2 * 4 = 5*9.8

w= 3.8858 rad/sec

when the cord goes slack, tension will be zero in the string

so only 2 forces will be acting on it mg and mw^2*r

to balance the net downward force, angle should be greater then 90 degree

let phi = theta - 90

mg= 0.5*mw^2 * r *cos(phi)

9.8 = 3.8858^2 * cos(phi)

cos(phi) = 0.649

phi = 49.53 degree

so theta = 90+ 49.53

= 139.53 degree

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