Force on roll of paper - length of paper that unrolls

[physicsss]

The radius of the roll of paper is 7.6 cm and its moment of inertia is I = 2.9 10^-3 kg · m2. A force of 3.2 N is exerted on the end of the roll for 1.3 s, but the paper does not tear so it begins to unroll. A constant friction torque of 0.11 m · N is exerted on the roll which gradually brings it to a stop. Assume that the paper's thickness is negligible.


(a) Calculate the length of paper that unrolls during the time that the force is applied (1.3 s).

(b) Calculate the length of paper that unrolls from the time the force ends to the time when the roll has stopped moving.

 

[Sirus]

Have you tried it?

Consider these formulas:
\sum{\tau}=I\alpha
\tau=Fl
where \alpha is angular acceleration and l is length of the lever arm.

 

[wais]

a) To calculate the length of paper that unrolls during the 1.3 seconds of force application, we can use the equation for rotational motion: τ = Iα, where τ is the torque, I is the moment of inertia, and α is the angular acceleration. We can rearrange this equation to solve for α: α = τ/I. The force of 3.2 N can be converted to torque by multiplying it by the radius of the roll (0.076 m). Plugging in the values, we get α = (3.2 N)(0.076 m)/(2.9 x 10^-3 kg · m^2) = 0.084 rad/s^2.

Next, we can use the equation for angular displacement to calculate the length of paper that unrolls: θ = ω0t + 1/2αt^2, where ω0 is the initial angular velocity (which we can assume to be 0 since the roll is not initially moving) and t is the time. Plugging in the values, we get θ = 0 + 1/2(0.084 rad/s^2)(1.3 s)^2 = 0.072 rad.

Since 1 revolution = 2π radians, we can convert this to the length of paper unrolled by multiplying by the radius of the roll: (0.072 rad)(0.076 m) = 0.00547 m or 5.47 cm. Therefore, the length of paper that unrolls during the 1.3 seconds of force application is approximately 5.47 cm.

b) To calculate the length of paper that unrolls after the force ends, we can use the equation τ = Iα again. However, this time, we know that the force of friction (0.11 m · N) is acting on the roll, so we can subtract this from the torque: τ = Iα - Ff, where Ff is the force of friction. We can rearrange this equation to solve for α: α = (τ + Ff)/I.

The torque can be calculated using the equation for rotational motion: τ = Iα, where τ is the torque, I is the moment of inertia, and α is the angular acceleration. We can plug in the values and solve for α: α = (0.11 m · N)/(2.