Respuesta :
Answer:
Normal force will be equal to 8.945 N
Explanation:
We have given mass of the cylinder m = 8.15 kg
Diameter d = 15 cm
So radius [tex]r=\frac{d}{2}=\frac{15}{2}=7.5cm = 0.075 m[/tex]
Initial angular velocity [tex]\omega _i=240rpm=\frac{2\times 3.14\times 240}{60}=25.12rad/sec[/tex]
As the cylinder finally comes to rest so final angular velocity [tex]\omega _f=0rad/sec[/tex]
Before coming to rest cylinder covers a distance of [tex]\Theta =5.20revolution=5.20\times 2\times 3.14=32.656rad[/tex]
From third equation of motion [tex]\omega _f^2=\omega_i^2+2\alpha \Theta[/tex]
[tex]0^2=25.12^2-2\times \alpha \times32.656[/tex]
[tex]\alpha =9.66rad/sec^2[/tex]
Coefficient of kinetic friction [tex]\mu _k=0.33[/tex]
Moment of inertia of the solid cylinder [tex]I=\frac{1}{2}mr^2[/tex]
We know that [tex]\tau =I\alpha[/tex]
[tex]F\times r =(\frac{1}{2}mr^2)\times \alpha[/tex]
[tex]F =(\frac{1}{2}mr)\times \alpha[/tex]
[tex]F=\frac{1}{2}\times 8.15\times 0.075\times 9.66=2.952N[/tex]
So normal force will be equal to [tex]N=\frac{F}{\mu _k}=\frac{2.952}{0.33}=8.945N[/tex]
The normal force which must be applied to bring the cylinder to rest is 8.955 Newton.
Given the following data:
- Mass = 8.15 kg
- Diameter = 15 cm to m = 0.15 m
- Angular velocity = 240 rpm
- Coefficient of kinetic friction = 0.330
- Angle = 5.20 revs.
Conversion:
Angular velocity = 240 rpm = [tex]240 \times \frac{2\pi}{60} = 25.12\;rads[/tex]
Angle = 5.20 revs to rad = [tex]5.20 \times 2 \times \pi = 32.66 \;rad[/tex]
Radius = [tex]\frac{Diameter}{2} = \frac{0.15}{2} = 0.075\;m[/tex]
First of al, we would calculate the angular acceleration of the cylinder by using the third equation of rotational motion (kinematics):
[tex]\omega_f^2 = \omega_i^2 - 2\alpha \theta\\\\\alpha = \frac{\omega_f^2 + \omega_i^2}{2\theta}[/tex]
Substituting the given parameters into the formula, we have;
[tex]\alpha =\frac{0^2\; +\;25.12^2}{2 \times32.66} \\\\\alpha =\frac{631.0144}{65.32} \\\\\alpha =9.66\;rad/s^2[/tex]
To find the normal force, we would apply the law of conservation of momentum:
Torque = Angular torque
[tex]Force \times radius = moment \;of\;inertia \times angular \;acceleration\\\\Fr = I\alpha[/tex]....equation 1.
For a solid cylinder, the moment of inertia is given by:
[tex]I = \frac{1}{2} mr^2[/tex] ....equation 2.
Substituting eqn. 2 into eqn. 1, we have:
[tex]Fr = (\frac{1}{2} mr^2)\alpha \\\\2F=mr\alpha \\\\F=\frac{mr\alpha}{2} \\\\F = \frac{8.15 \times 0.075 \times 9.66}{2} \\\\F=\frac{5.91}{2}[/tex]
Force = 2.955 Newton
For the normal force:
[tex]F_N = \frac{F}{u} \\\\F_N = \frac{2.955}{0.330}[/tex]
Normal force = 8.955 Newton.
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