Answer:
[tex]\mu_s=\frac{1}{3}\tan \theta[/tex]
Explanation:
Let the minimum coefficient of static friction be [tex]\mu_s[/tex].
Given:
Mass of the cylinder = [tex]M[/tex]
Radius of the cylinder = [tex]R[/tex]
Length of the cylinder = [tex]L[/tex]
Angle of inclination = [tex]\theta[/tex]
Initial velocity of the cylinder (Released from rest) = 0
Since, the cylinder is translating and rolling down the incline, it has both translational and rotational motion. So, we need to consider the effect of moment of Inertia also.
We know that, for a rolling object, torque acting on it is given as the product of moment of inertia and its angular acceleration. So,
[tex]\tau =I\alpha[/tex]
Now, angular acceleration is given as:
[tex]\alpha = \frac{a}{R}\\Where, a\rightarrow \textrm{linear acceleration of the cylinder}[/tex]
Also, moment of inertia for a cylinder is given as:
[tex]I=\frac{MR^2}{2}[/tex]
Therefore, the torque acting on the cylinder can be rewritten as:
[tex]\tau = \frac{MR^2}{2}\times \frac{a}{R}=\frac{MRa}{2}------ 1[/tex]
Consider the free body diagram of the cylinder on the incline. The forces acting along the incline are [tex]mg\sin \theta\ and\ f[/tex]. The net force acting along the incline is given as:
[tex]F_{net}=Mg\sin \theta-f\\But,\ f=\mu_s N\\So, F_{net}=Mg\sin \theta -\mu_s N-------- 2[/tex]
Now, consider the forces acting perpendicular to the incline. As there is no motion in the perpendicular direction, net force is zero.
So, [tex]N=Mg\cos \theta[/tex]
Plugging in [tex]N=Mg\cos \theta[/tex] in equation (2), we get
[tex]F_{net}=Mg\sin \theta -\mu_s Mg\cos \theta\\F_{net}=Mg(\sin \theta-\mu_s \cos \theta)--------------3[/tex]
Now, as per Newton's second law,
[tex]F_{net}=Ma\\Mg(\sin \theta-\mu_s \cos \theta)=Ma\\\therefore a=g(\sin \theta-\mu_s \cos \theta)------4[/tex]
Now, torque acting on the cylinder is provided by the frictional force and is given as the product of frictional force and radius of the cylinder.
[tex]\tau=fR\\\frac{MRa}{2}=\mu_sMg\cos \theta\times R\\\\a=2\times \mu_sg\cos \theta\\\\But, a=g(\sin \theta-\mu_s \cos \theta)\\\\\therefore g(\sin \theta-\mu_s \cos \theta)=2\times \mu_sg\cos \theta\\\\\sin \theta-\mu_s \cos \theta=2\mu_s\cos \theta\\\\\sin \theta=2\mu_s\cos \theta+\mu_s\cos \theta\\\\\sin \theta=3\mu_s \cos \theta\\\\\mu_s=\frac{\sin \theta}{3\cos \theta}\\\\\mu_s=\frac{1}{3}\tan \theta............(\because \frac{\sin \theta}{\cos \theta}=\tan \theta)[/tex]
Therefore, the minimum coefficient of static friction needed for the cylinder to roll down without slipping is given as:
[tex]\mu_s=\frac{1}{3}\tan \theta[/tex]
The minimum coefficient of static friction that needed for cylinder to roll down without slipping is [tex]\mu_s= \frac{tan\theta}{3}[/tex]
A cylinder of radius R, length L, and mass M is released from rest on a slope inclined at angle θ. It is oriented to roll straight down the slope. If the slope were frictionless, the cylinder would slide down the slope without rotating. What minimum coefficient of static friction is needed for the cylinder to roll down without slipping?
Given: radius R, length L, angle θ, and mass M
We need to calcuate the minimum static friction coefficient. It is useful so the cylinder will roll without slipping down the incline. The cylinder is also released from rest. As the cylinder is rolling, we have to consider the moment of inertia. Rolling of cylinder is happened due to the friction force
By applying Newton law of motion
[tex]F = M a\\\tau = I \alpha\\\tau = I \frac{a}{R} \\\tau = \frac{1}{2} M R^2 \frac{a}{R}[/tex]
From diagram
[tex]Mg sin\theta - f_{fr} = Ma\\f_{fr} = \mu_s N\\f_{fr} = \mu_s Mg cos \theta\\a = g sin \theta - \mu_s cos \theta[/tex]
Then also
[tex]\tau = f_{fr} R\\f_{fr} = \frac{Ma}{2} \\\mu_s Mg cos \theta = \frac{Mg (sin\theta - \mu_s cos \theta)}{2} \\\frac{3}{2} \mu_s cos\theta = \frac{sin\theta}{2}\\ \mu_s = \frac{tan\theta}{3}[/tex]
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