solve the questions in bold ? One may illustrate the principles that you will ex

Question

solve the questions in bold ?

One may illustrate the principles that you will examine in this lab by studying a falling bucket.


A frictionless pulley in the shape of a solid cylinder of unknown mass (M = ?) and radius r = 0.150 m is used to draw water from a well. A bucket of mass m = 1.50 kg is attached to a cord wrapped around the cylinder. The bucket starts from rest at the top of the well and falls for t = 2.2 s before hitting the water h = 6.00 m below the top of the well. Neglect the mass of the cord.

(a) What is the linear acceleration of the falling bucket?
Hint: h = at2/2  a = 2h/t2
Linear acceleration = m/s2

(b) What is the angular acceleration of the rotating pulley? rad/s2

(c) What is the tension in the cord? N

(d) What is the torque that is applied to the pulley due to the cord? (use the tension from the previous question) N*m

(e) Using the torque and the angular acceleration, find the moment of inertia of the pulley. kg*m2

(f) Using the moment of inertia, find the mass of the pulley. kg

(g) What is the change in the potential energy of the bucket? J

(h) What is the linear velocity of the bucket when it hits the water? (use the previously calculated linear acceleration) m/s

(i) What is the angular velocity of the pulley when the bucket hits the water? (use the relation between the angular velocity and the linear velocity) rad/s

(j) What is the angular momentum of the pulley when the bucket hits the water? (use the moment of inertia and the angular velocity of the pulley) kg*m2/s

(k) What is the kinetic energy of the bucket when it hits the water? J

(l) What is the kinetic energy of the pulley when the bucket hits the water? J

(m) What is the total kinetic energy of the falling bucket and the rotating pulley when the bucket hits the water? J

(n) Is the total kinetic energy KE smaller, equal to or larger than the potential energy change ?PE? Ignore the error at the last digit that is due to the number rounding!

Explanation / Answer

a)
h = 0.5*a*t^2

a = 2*h/t^2 = 2*6/2.2^2 = 2.48 m/s^2

b) alfa = a_tan/r = 2.48/0.15 = 16.53 rad/s^2

c) T = m*g - m*a = 1.5*9.8 - 1.5*2.48 = 10.98 N

d) Torque, T = tension*r = 10.98*0.15 = 1.647 N.m

e) Torque = I*alfa

==> I= torque/alfa = 0.09964 kg.m^2

f) I = 0.5*M*r^2

M = 2*I/r^2 = 8.86 kg.

g)delta U = -m*g*h = -1.5*9.8*6 = -88.2 J

h)v = u + a*t = 0 + 2.48*2.2 = 5.456 m/s

i) w = v/r = 36.37 rad/s

j) L = I*w = 3.62 kg.m^2/s

k) KE = 0.5*m*v^2 = 22.33 J

l)KE = 0.5*I*w^2 = 65.9

m) KE(total) = 88.23 J

n)loss of potential energy = increase in kinetic energy

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