3. 15 points SerCP11 7.P.054 My Notes Ask Your Teac A 0.440-kg pendulum bob pass

Question

3. 15 points SerCP11 7.P.054 My Notes Ask Your Teac A 0.440-kg pendulum bob passes through the lowest part of its path at a speed of 3.06 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 Heln?

Explanation / Answer

(a) r=78 cm= 0.78 m

T- mg cos theta= mv2/r

at lowsest point, theta= 0

T= mv2/r + mg

T=0.44(3.06)2/0.78 + 0.44 (9.8)

T=9.59 N

(b)angle with vertical at highest point= 180 degrees

(c)let Vh= velocity at highest point, we use the conservation of energy to find v

1/2 mVh2+ mgh= 1/2mv2

Vh2= v2- 2 gh

h= 2r= 2(0.78)=1.56m

Vh2= (3.06)2 - 2 *9.8 * 1.56

Vh2=-21.2

Vh comes out to be imaginary. that means the bob will slack before reaching highest point. so tension at top= 0

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