The figure shows a pendulum of length L = 1.2 m. Its bob (which effectively has

Question

The figure shows a pendulum of length L = 1.2 m. Its bob (which effectively has all the mass) has speed vo when the cord makes an angle eo = 46" with the vertical. (a) What is the speed of the bob when it is in its lowest position if vo-6.6 m/s? What is the least value that vo can have if the pendulum is to swing down and then up (b) to a horizontal position, and (c) to a vertical position with the cord remaining straight? (d) Do the answers to (b) and (c) increase, decrease, or remain the same if 0 is increased by a few degrees?

Explanation / Answer

a) we have (K + U) 1 = K2

So 1/2*m*v1^2 + m*g*L*(1-cos(46)) = 1/2*m*v2^2

So v2 = sqrt(v1^2 + 2*g*L*(1 - cos(46)) = sqrt(6.6^2 + 2*9.81*1.2*(1 - cos(46)) = 6.03 m/s

b)Now (K + U)1 = U2

so 1/2*m*v1^2 + m*g*L*(1-cos(46)) = m*g*L

so v1 = sqrt(2*(g*L*(1 - (1-cos(46))) = sqrt(2*(9.81*1.2*cos(46))) = 4.04 m/s

c) Now (K + U)1 = K2 + U2

There is a K2 if the cord is to remain straight

We have T + m*g = m*v^2/L f T = 0 then v^2 = g*L

so 1/2*m*v1^2 + m*g*L*(1-cos(46)) = m*g*2*L + 1/2*m*g*L

so v1^2 = 4*g*L - 2*g*L*(1 - cos(46)) + g*L = 4*9.81*1.2 - 2*9.81*1.2*(1- cos(46)) + 9.81*1.2 = 51.67.

So v1 = sqrt(51.67) = 7.19 m/s

d) b - decrease

c - decrease

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