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erastova [34]
3 years ago
7

A 15 kg bucket of water is suspended by a very light ropewrapped around a solid uniform cylinder 0.300 m in diamter withmass 12.

0 kg. The cylinder pivots on a frictionless axle throughits center. The bucket is relased from rest at the top of awell and falls 10 m to the water.
a) What is the tension in the rope while the bucket isfalling?
b) what speed does the bucket strike the water?
c) What is the time of fall?
d) While the bucket is fallin, what is the force exterded onthe cylinder by the axle?
Physics
1 answer:
matrenka [14]3 years ago
5 0

Answer:

Part a)

T = 42 N

Part b)

v_f = 11.8 m/s

Part c)

t = 1.7 s

Part d)

F = 159.7 N

Explanation:

Part a)

While bucket is falling downwards we have force equation of the bucket given as

mg - T = ma

for uniform cylinder we will have

TR = I\alpha

so we have

T = \frac{1}{2}MR^2(\frac{a}{R^2})

T = \frac{1}{2}Ma

now we have

mg = (\frac{M}{2} + m)a

a = \frac{mg}{(\frac{M}{2} + m)}

a = \frac{15 \times 9.81}{(6 + 15)}

a = 7 m/s^2

now we have

T = \frac{12 \times 7}{2}

T = 42 N

Part b)

speed of the bucket can be found using kinematics

so we have

v_f^2 - v_i^2 = 2 a d

v_f^2 - 0 = 2(7)(10)

v_f = 11.8 m/s

Part c)

now in order to find the time of fall we can use another equation

v_f - v_i = at

11.8 - 0 = 7 t

t = 1.7 s

Part d)

as we know that cylinder is at rest and not moving downwards

so here we can use force balance

F = T + Mg

F = 42 + (12 \times 9.81)

F = 159.7 N

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