3. + -11.66 points SerCP117.P.054. A 0.420-kg pendulum bob passes through the lo

Question

3. + -11.66 points SerCP117.P.054. A 0.420-kg pendulum bob passes through the lowest part of its path at a speed of 3.21 m/s. (a) What is the magnitude of the tension in the pendulum cable at this point if the pendulum is 78.0 cm long? (b) When the pendulum reaches its highest point, what angle does the cable make with the vertical? (Enter your answer to at least one decimal place.) (c) What is the magnitude of the tension in the pendulum cable when the pendulum reaches its highest point? Need Help? Read It MacBook Air

Explanation / Answer

given

m = 0.42 kg
v = 3.21 m/s
a) L = 78 cm = 0.78 m

Tension in the string, T = m*g + m*v^2/L

= 0.42*9.8 + 0.42*3.21^2/0.78

= 9.66 N

b) maximum height reached, h = v^2/(2*g)

= 3.21^2/(2*9.8)

= 0.5257 m

let theta is the angle made by the thread with vertical.

h = L - L*cos(theta)

L*cos(theta) = L - h

cos(theta) = 1 - h/L

cos(theta) = 1 - 0.5257/0.78

theta = cos^-1(0.326)

= 71.0 degrees

c) T*cos(71) = m*g

T = m*g/cos(71)

= 0.42*9.8/cos(71)

= 12.6 N

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