A small rock with mass .12kg is fastened to a massless string with length 0.80 m

Question

A small rock with mass .12kg is fastened to a
massless string with length 0.80 m to form a pendulum. The pendulumis
swinging so as to make a maximum angle of 45* with vertical.Air
resistance is negligible.



(a) What is the speed of the rock when the string passes throughthe vertical position?



(b) What is the tension in the string when it makes an angle of 45*with a vertical?



(c) What is the tension in the string as it passes through thevertical?







All steps and answers would be greatly appreciated! Thank you

Explanation / Answer

(b) At = 45, v = 0 for the rock, as it is at itsmaximum height and just returning. The force acting on the rock are, its weight, mg and thetension in the string, T. Resolving mg into perpendicular components, along the stringand perpendicular to it, mg cos acts out ward and tension T acts inward, in theopposite direction. Since v = o, the centripetal force Fc = m.v2/r = 0;the net force is zero. Hence T - mg cos = 0
             T = mg cos
                = 0.12 * 9.8 * cos 45
                = 0.8315 N
(c) Tension in the string when it passes through the mean positionis
             T = mg + mv2 / l
                = m g +   m * 2gl / l
                = 3 m g
                = 3 * 0.12 * 9.8
                = 3.528 N

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