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Chapter 5: Problem 595

A Pulley of radius \(2 \mathrm{~m}\) is rotated about its axis by a force \(F=\left(20 t-5 t^{2}\right) N\) where \(t\) is in sec applied tangentially. If the moment of inertia of the Pulley about its axis of rotation is $10 \mathrm{KgM}^{2}$, the number of rotations made by the pulley before its direction of motion is reversed is : \(\\{\mathrm{A}\\}\) more than 3 but less then 6 \(\\{\mathrm{B}\\}\) more than 6 but less then 9 \(\\{\mathrm{C}\\}\) more than 9 \\{D \\} Less then 3

Short Answer

Expert verified
The number of rotations made by the pulley before its direction of motion is reversed is more than 9 rotations (Option C).

Step by step solution

01

Calculate the torque on the pulley

In order to find the torque on the pulley, we need to multiply the force applied by the radius of the pulley. The force applied on the pulley is given by \(F(t) = (20t - 5t^2) N\). Multiply this force by the radius of the pulley, which is 2 meters: \[τ(t) = r \cdot F(t) = 2 \cdot (20t - 5t^2) = 40t - 10t^2 \mathrm{~N m}\]
02

Calculate the angular acceleration of the pulley

Now that we have the torque on the pulley, we can use the relationship between torque and moment of inertia, \(\tau = I \cdot \alpha\), where \(\tau\) is the torque, \(I\) is the moment of inertia, and \(\alpha\) is the angular acceleration. Solve for \(\alpha\): \[ \alpha(t) = \frac{\tau(t)}{I} = \frac{40t - 10t^2}{10} = 4t - t^2 \mathrm{~rad/s^2}\]
03

Integrate the angular acceleration function

Next, integrate the angular acceleration function with respect to time to find the angular velocity function, \(\omega(t)\). Don't forget to add an integration constant, \(C_1\): \[ \omega(t) = \int \alpha(t) dt = \int (4t - t^2)dt = 2t^2 - \frac{1}{3}t^3 + C_1 \mathrm{~rad/s}\]
04

Determine when the angular velocity becomes zero

To find when the direction of motion is reversed, we need to find when the angular velocity becomes zero. Solving for \(t\) in \(\omega(t) = 0\): \[ 2t^2 - \frac{1}{3}t^3 + C_1 = 0\] Assuming the pulley starts from rest, \(C_1 = 0\). So, solve the equation for \(t\): \[ 2t^2 - \frac{1}{3}t^3 = 0\] \[\frac{1}{3}t^3 = 2t^2\] \[ t = \frac{6}{1}\] So, the pulley reverses its direction at \(t = 6s\).
05

Calculate the total number of rotations before reversal

Finally, to find the total number of rotations, we need to find the total angle covered by the pulley by integrating the angular velocity function: \[θ(t) = \int \omega(t) dt = \int_0^6 (2t^2 - \frac{1}{3}t^3) dt = \left[\frac{2}{3}t^3 - \frac{1}{12}t^4\right]_0^6\] \[ θ(6) = \frac{2}{3}(6)^3 - \frac{1}{12}(6)^4 = 72\pi\] Now, divide this angle by a full rotation, which is \(2\pi\) radians: \[ \text{number of rotations} = \frac{72\pi}{2\pi} = 36\] The number of rotations made by the pulley before its direction of motion is reversed is 36. Which corresponds to option C: More than 9 rotations.

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