Problem 1

A simple pendulum consists of a small bob of mass

m=0.50

kg on a light inextensible string of length

L=1.25

m attached to a fixed support. The string is displaced to one side until it makes

60^\circ

with the vertical and is released from rest. Neglect resistance and take

g=10

m/s

^2

. (a) Find the bob's height above the lowest point just before release. (b) Find its speed through the lowest point. (c) Find the string tension there. (d) Find the angle to the vertical where kinetic energy equals potential energy above the lowest point.

problems.proof.stepOf

problems.proof.prevStep problems.proof.nextStep

T-mg=\frac{mv_0^2}{L}=mg\Rightarrow T=2mg=10\ \mathrm N

problems.proof.analysis

At the lowest point the velocity is horizontal and the centripetal acceleration points upward. Projecting Newton's second law vertically gives $T-mg=mv_0^2/L$ ; since $v_0^2=gL$ , $T=2mg$ .

problems.proof.pitfall. Do not set

T=mg

at the lowest point: the bob has upward centripetal acceleration, so tension must exceed weight.