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