A constant horizontal force of 150 n is applied to a lawn...

A constant horizontal force of 150 N is applied to a lawn roller in the form of a uniform solid cylinder of radius 0.4 m and mass 13 kg . If the roller rolls without slipping, find the acceleration of the center of mass. The acceleration of gravity is 9.8 m/s^2. Answer in units of m/s^2. Then, find the minimum coefficient of friction necessary to prevent slipping.

Transcript

00:01 According to newton's second law of motion, net force on an object is equal to the product of its mass and acceleration.

00:06 In this problem, a lawnmower, this is in a shape of a uniform solid cylinder that has a mass of 13 kilograms and radius of 0 .4 meters is being acted upon by a concept horizontal force of 15 neutons.

00:19 If the rollers rolls without sleeping, we wish to find the acceleration of its center of mass and the coefficient of friction necessary to prevent it from sleeping.

00:29 So the net force on the wheel, or in this case the lawnmower, is equal to the difference between the applied force and the friction.

00:43 It's the difference between these two forces as the two forces looking at the pre -body diagram are opposing each other.

00:50 So equals mass times acceleration.

00:53 Now we note that friction is applied between the surface of between the surface and the, the surface and the, the cylinder itself so that this force would actually cause the the we call this the wheel let's just call this as wheel to rotate about the fixed point which is its center so this force of friction will cause torque and we quantify that torque to be equal to the frictional force times the distance of the friction from the rotation axis which corresponds to the radius and by definition we also note that torque is equal to the moment of inertia of the wheel times the angular acceleration so what we're trying to do here is that we will express the frictional force in terms of i alpha and the r so i is the moment of inertia which the value or which formula depends on the shape in this case cylinder alpha is just let's just take alpha as alpha here first and then radius is r now for cylinder your i is equal to one half times the mass times the square of the radius now also alpha in terms of the moment or in terms of the cent a assertion of the center of mass is a divided by r so with that f is equal to we have i which is one half times m r squared multiplied by the alpha in terms of acceleration of center of mass is a divided by r divided by the radius.

02:48 Simplifying this, use as first is we can rewrite this as mr squared a divided by 2r squared.

03:02 Or clearly this becomes ma over 2.

03:09 Now going back to our net force equation where we have f minus the applied force minus the friction, which equals to mass times acceleration.

03:20 So we're using capital m here for the mass.

03:23 Then we have f minus m a over 2 equals m times a...

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