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

∑ F_x  = 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 )