Physics: rotational motion

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A 0.480-kg pendulum bob passes through the lowest part of its path at a speed of 3.26 m/s.

(a) What is the magnitude of the tension in the pendulum cable at this point if the pendulum is 73.0 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? 

Jun 30th, 2015

Thank you for the opportunity to help you with your question!

T = mg + mv²/r 
T = m(g + v²/r) 
T = (0.480 kg)[(9.81 m/s²) + (3.26 m/s)²/(0.800 m)] 
T = 11.8 N 

(1/2)mv² = mgy 
(1/2)v² = gy 
(1/2)(3.26 m/s)² = (9.81 m/s²)(y) 
y = 0.6244 m 

cosθ = (L - y)/L 
cosθ = (0.800 - 0.6244 m) / (0.800 m) 
θ = 77.3° 

T = mgcosθ 
T = (0.470 kg)(9.81 m/s²)cos77.3° 
T = 1.01 N

note.....made recalculation........error may be occure

Please let me know if you nea case for torture by michale levined any clarification. I'm always happy to answer your questions.
Jun 30th, 2015

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