Question

A lawn roller in the form of a thin-walled, hollow cylinder with mass M is pulled horizontally with a constant horizontal force F applied by a handle attached to the axle. If it rolls without slipping, find the acceleration and the friction force.

Answer 1

Answer:

$a_{cm}=\frac{F}{2M}\\\\F_{fr}=\frac{F}{2}$

Explanation:

Given the mass as M, the rotational inertia of the mower is;

$I_{cm}=MR^2$

-The roller doesn't slip while rolling;

$v_{cm}=wR, a_{cm}=\alpha R$

$\sum F_x=Ma_x, -F+F_{fr}=-Ma_{cm}\\\\a_{cm}=\frac{F-Fr}{M} \ \ \ \ \ \ \ \ eqtn1\\\\\sum \tau_{cm}=I_{cm}\alpha, F_{fr}(R)}=(MR^2)(\frac{a_{cm}}{R}), ->F_{fr}=Ma_c_m\ \ \ \ \ \ \ eqtn2\\\\a_{cm}=\frac{F-Ma_{cm}}{M}, ->F=2Ma_{cm}\\\\a_{cm}=\frac{F}{2M}\\\\\\\therefore F_{fr}=M(\frac{F}{2M})\\\\F_{fr}=\frac{F}{2}$

Answer 2

The acceleration will be "a = $\frac{F}{2m}$" and the friction force will be "$F_{f} = \frac{F}{2}$".

Friction force

According to the question,

As whenever the linear velocity rises, additional friction force ($F_f$) opposes the externally force applied (F), causing the angular velocity just to expand.

Newton's law would be required including both forces as well as the torques.

Now,

→ F - $F_f$ = $M_a$

    $F_f$ R = Iα

          a = Rα

When all mass on the surface,

I = $M_r$²

F - $\frac{I \alpha}{R}$ = $M_a$

F - $M_a$ = $M_a$

hence,

The acceleration, a = $\frac{F}{2M}$

The friction force, $F_f$ = $M_a$

                                   = $\frac{F}{2}$

Thus the response above is correct.

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