The pulley shown in the figure below has a moment of inertia, I, and a radius, R
ID: 1477396 • Letter: T
Question
The pulley shown in the figure below has a moment of inertia, I, and a radius, R. The masses of the two objects are, mA and mB. There is friction between object A and the horizontal surface with the coefficient of kinetic friction, µ K .
(a) Draw the sketch below onto your homework sheet and show all forces acting on each of the two objects as well as on the pulley. Show a free-body-diagram for each of the two objects and set up Newton’s 2nd Law (in vector form). Decide what you want your positive directions to be and work out a relationship between the tension, mass, and acceleration for each body. The system is accelerating in clockwise direction. Now focus on the pulley and in a separate sketch, mark all forces acting on the pulley and determine their torques with respect to the axis of rotation of the pulley. Argue whether the angular acceleration of the pulley is positive or negative. Using Newton’s 2nd Law for the rotational motion, set up a relationship between the tensions in the string and the angular acceleration.
(b) Find an expression for the acceleration of the system depending only on the given quantities, i.e. the two masses, I, R, the coefficient of kinetic friction, and g.
(c) Find an expression for the speed of block B after it fell through a distance d. This expression should again depend on the above quantities and on d. (d) Calculate the acceleration, tensions in the string andangular acceleration if I = 0.0025kgm 2 , R =10cm,µ K = 0.2,mA = mB = 2kg .
(d) Find an expression for the speed of block B from problem 6 after it fell through a height d, using energy methods. Which method do you find easier?
Explanation / Answer
for A
TA - u*mA*g = mA*a
TA = mA*a + u*mA*g ......(1)
for B
mB*g - TB = mB*a
TB = mB*g - mB*a............(2)
for pulley
(TB - TA )*R= I*alpha
(mB*g - mB*a - mA*a - u*mA*g)*R = I*a/R
((2*9.8)-(2*a)-(2*a)-(0.2*2*9.8))*0.1=0.0025*a/0.1
a = 3.68 m/s^2
---------------------
for B
v = sqrt(2*a*d) = sqrt(2*3.68*6) = 6.645 m/s <<-----------answer
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