The radius of the roll of paper shown in the figure is 7.6 and its moment of ine
ID: 1978213 • Letter: T
Question
The radius of the roll of paper shown in the figure is 7.6 and its moment of inertia is = 3.1×10-3 . A force of 2.3 is exerted on the end of the roll for 1.0 , but the paper does not tear so it begins to unroll. A constant friction torque of 0.11 is exerted on the roll which gradually brings it to a stop.A) Assuming that the paper's thickness is negligible, calculate the length of paper that unrolls during the time that the force is applied (1.0 ).(in meters)
B) Assuming that the paper's thickness is negligible, calculate the length of paper that unrolls from the time the force ends to the time when the roll has stopped moving. (in meters)
Explanation / Answer
solved a similar question already with different numbers.. i put the question and the answer.. hope it helps!! :)The radius of the roll of paper is 0.08m and its moment of inertia is I = 3.5×10-3 kg*m^2. A force of 2.3 N is exerted on the end of the roll for 1.0 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. a) Assuming that the paper's thickness is negligible, calculate the length of paper that unrolls during the time that the force is applied (1.0 s). (This answer is around 1 or 2 m) b) Assuming that the paper's thickness is negligible, calculate the length of paper that unrolls from the time the force ends to the time when the roll has stopped moving. (This answer is around (0.5 or 1.5m) ANSWER: When rotations come into play the formula F = m*a changes into tau = I * alpha with tau the torque (= F * r) in Nm and alpha the angular acceleration in radians/second^2. Neglecting the paper's thickness means that it is assumed that the moment of inertia I does not change when you roll off some paper. alpha = tau / I = F*r / I With a constant angular acceleration, the angular velocity omega in radians/second is omega(t) = omega(0) + alpha*t And the angular displacement theta in radians is theta(t) = theta(0) + omega(0)*t + 1/2 * alpha * t^2 a) At t = 0: theta(0) = 0 and omega(0) = 0. The torque is a combination of the pulling force and the friction: tau = F*r - 0.11 = 2.3*0.08 - 0.11 = 0.074 Nm alpha = tau / I = 0.074 / 3.5*10^-3 = 21.14 rad/s^2 theta(1.0) = 1/2 * alpha * (1.0)^2 = 10.57 rad One time around the roll is 2*pi radians. That is an unrolling distance of 2*pi*r. That means that an angular displacement of 1 rad is an unrolling distance of 1*r meters (definition of a radian). So the unrolling distance x in meters is x = 10.57 * r = 10.57 * 0.08 = 0.85 m b) Now only the friction torque is left: tau = -0.11 Nm First you have to know how long it takes to stop the roll. The end-speed of question a) is omega(1) = alpha*1 = 21.14 rad/s This is the start-speed of this part: omega(0) = 21.14 rad/s. The time at which omega(t) is zero under the given torque is omega(t) = omega(0) + alpha*t 0 = omega(0) + tau/I *t_end t_end = -omega(0)*I/tau = -21.14 * 3.5*10^-3 / (-0.11) = 0.6726 s The angular displacement in this time is theta(t) = theta(0) + omega(0)*t + 1/2 * alpha * t^2 theta(t_end) = 0 + 21.14*0.6726 + 1/2 * (-0.11 / 3.5*10^-3) * 0.6726^2 = 14.22 - 7.109 = 7.111 rad And the unrolling distance is