Find out the friction force:
The cord passes over a massless and frictionless pulley, carrying a mass M1 at one end and wrapped around a cylinder of mass M2 which rolls on a horizontal plane. What is the acceleration of mass M1?
Solution
We should keep in mind that acceleration a1 of mass M1 is not equal to the acceleration a of the mass centre of the cylinder.
Considering dynamic equilibrium, for cylinder we may write
∑ Fx = 0, ∴ T - M 2 a¯ - F = 0 -------- (1)
∑ M o = 0, T r + F . r - I α = 0 -------- (2)
and for mass M1, we have
∑ Fy = 0, T - M1 g + M1 a1 = 0 -------- (3)
and, we have,
a¯ = r α
and, a1 = 2 r α
i.e., a1 = 2 a¯
and
I =( ½) M2 r 2 , a¯ = r α
∴ From Eq. (1) above, we obtain
T - F = M 2 a¯
From Eq. (2)
(T + F) r = (½) M 2 r 2 α
T + F = (1/2)M 2 r α
T + F = 1 M2 a¯
∴ 2T = M 2 a¯ + (1 /2)M2 a
∴ T = (¾) M2 a¯ ---------- (4)
Figure 7.17
From Eq. (3), we get
T = M 1 g - M1 a1
∴ From Eq. (4),
T = (¾) M2a=M1g-M1a1
But a¯ = (1/2)a1
∴ 3/8 M2a1 = M1 g - M1 a1
∴ a1((3/8) M 2 + M1 ) = M1 g
∴ a1 = M1 g/(3/8) M 2 + M1 )