The 40 kg roll of paper rests along the wall where the coefficient of kinetic friction is µk = 0.2. If a vertical force of P = 40 N is applied to the paper, determine the angular velocity of the roll when t = 6 s starting from rest. Neglect the mass of the unraveled paper and take the radius of gyration of the spool about the axle O to be kO = 80 mm.

The answer is w = 36.36 rad/s

Explanation:

Solution

Recall that,

Roll of paper rests along the wall it's coefficient of kinetic friction is µk= 0.2

Vertical force P = 40 N

The angular velocity  of the roll when t =6

Then,

The equilibrium  force on vertical direction  is given as

∑Fy = ma

T = (12/13) + P  + 0.2Na = mg

0.923T + 0.2Na = 352.4 this is the equation (1)

Now,

The equilibrium force on horizontal direction is given as

∑Fx =ma

Na=  T (5/13)

thus,

Na - 0.384T = 0 This is the equation (2)

Now,

∑M₀ = I₀α

= (0.2Na * 0.12)- (P * (0.12) = mk²α

so,

0.02µNa - µ0 * 0.12 = µ0 * 0.08²α

0.02µNa - µ*8 = 0.256α This is equation (3)

Now, we solve for equation 1 and 2

Na =  135.3µN, T = 352.47N

From the equation (3) we have,

α = 6.06 rad/s²

then,

at t= 65

The angular velocity , w= w₀ + αt

= 0 + 6.06 * 6

Therefore w = 36.36 rad/s

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