Two masses m1 = 2.5 kg, and m2 = 4.0 kg are connected by a light cord that passe

Question

Two masses m1 = 2.5 kg, and m2 = 4.0 kg are connected by a light cord that passes over a cylindrical pulley (I = MR2/2), where M = 1.2 kg, and R = 0.15m. The mass m2 slides on frictionless, horizontal surface. Find; a) The acceleration of the masses b) The angular acceleration of the cylinder c) The tensions in the strings

Two masses m1 = 2.5 kg, and m2 = 4.0 kg are connected by a light cord that passes over a cylindrical pulley (I = MR2/2), where M = 1.2 kg, and R = 0.15m. The mass m2 slides on frictionless, horizontal surface. Find; a) The acceleration of the masses b) The angular acceleration of the cylinder c) The tensions in the strings

Explanation / Answer

(a)

Let us make the following assumption:

T1 is the tension between m1 and the pulley

T2 is the tension between m2 and the pulley

BAlancing force on m1:

T1 = m1a---------------------------(*)

Balancing force on m2:

m2g - T2 = m2a

T2 = m2g - m2a--------------------------------(**)

Torque on pulley:

Torque= (T2 - T1)R ------------------(1)

Torque on pulley is also given as:

Torque= I* = I*a/R ---------------(2)

From (1) and (2), we have

(T2-T1)R = I*a/R
(T2 - T1) = Ia/R^2

From (*) and (**), we have

(m2g - m2a) - m1a = Ia/R^2
a(-m2 - m1 - I/R^2) = - m2g
a = m2g/(m1+m2+I/R^2)

=m2g/(m1+m2+0.5*Mp)  

a=4*9.8/(4+2.5+0.5*1.2)
= 5.521127

(b)

T1 = m1a = 2.5*5.521127 = 13.80282

T2 = m2g - m2a = 4*9.8-4*5.521127 = 17.11549

I = 1/2*mp*R^2

=0.5*1.2*0.15^2

= 0.0135

substituting the above value in following equation,we have

(T2 - T1) = I*

= 245.383

(c)

T1 = m1a = 2.5*5.521127 = 13.80282

T2 = m2g - m2a = 4*9.8-4*5.521127 = 17.11549

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