The radius of a roll of paper is 7.6 cm and its moment of inertia is 2.9 x 10^3 kg.m^2. A force...
Question:
The radius of a roll of paper is 7.6 cm and its moment of inertia is 2.9 x 10{eq}^3 {/eq} kg.m{eq}^2 {/eq}. 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 N.m 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.
Torque
The term torque refers to a physical quantity of vectorial character that depends on 3 things: the modulus of the applied force, the direction of the force, and the distance from the point of force application to the axis of rotation.
Answer and Explanation: 1
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View this answerVariables:
R is the radius of the cylinder (paper roll)
I is the moment of inertia
F is the applied force
{eq}\tau_F{/eq} is the torque of the...
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Chapter 7 / Lesson 5In this lesson, understand what is rotational inertia. See rotational inertia examples to learn how to find rotational inertia using the rotational inertia formula.
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