Two objects, of masses m1 = 455.0 g and m2 = 459.0 g, are connected by a string
ID: 1504991 • Letter: T
Question
Two objects, of masses m1 = 455.0 g and m2 = 459.0 g, are connected by a string of negligible mass that passes over a pulley with frictionless bearings. The pulley is a uniform 52.5-g disk with a radius of 3.96 cm. The string does not slip on the pulley.
(a) Find the accelerations of the objects. m/s2
(b) What is the tension in the string between the 455.0-g block and the pulley? (Round your answer to four decimal places.) N
-What is the tension in the string between the 459.0-g block and the pulley? (Round your answer to four decimal places.) N
-By how much do these tensions differ? (Round your answer to four decimal places.) N
(c) What would your answers be if you neglected the mass of the pulley?
-acceleration m/s2 -
-tension N
Explanation / Answer
given data
m1 = 455.0 g and m2 = 459.0 g
disk mass = 52.5 g
radius of 3.96 cm
The moment of inertia of the pulley is I = k*m*R² where k = ½. The value of k is all we need for this problem
(a) Find the accelerations of the objects
a = g[(2m2 + k)/(k+m1+m2) - 1]
a= 9.8*(((2*459*10^-3+0.5)/(0.5+455*10^-3+459*10^-3))-1) = 0.0277 m/s^2
(b) What is the tension in the string between the 455.0-g block and the pulley?
T1 = g*m1*m2*(2+k/m2)/(k+m1+m2)
T1 = 9.8*455*10^-3*459*10^-3*((2+0.5/(459*0^-3))/(0.5+455*10^-3+459*10^-3)) =2.894 N
What is the tension in the string between the 459.0-g block and the pulley?
T2 = g*m1*m2*(2+k/m1)/(k+m1+m2)
T2 = 9.8*455*10^-3*459*10^-3*((2+0.5/(455*0^-3))/(0.5+455*10^-3+459*10^-3))=2.894
(c) What would your answers be if you neglected the mass of the pulley?
For pulley mass = 0,
a = g*[1 - 2/(1+m2/m1)] and T = 2g/(1/m1+1/m2)
=2*9.8/((1/(455*10^-3))+(1/(459*10^-3))) =4.478 m/s^2
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