Respuesta :
Answer:
Explanation:
If '[tex]M[/tex]' be the mass of the roller and '[tex]R[/tex]' is the radius of its cylindrical shape, then moment of inertia '[tex]I[/tex]' can be written as,
[tex]I = MR^{2}[/tex]
If '[tex]F[/tex]' is the force by which the roller is pulling, '[tex]f[/tex]' is the frictional force and '[tex]a[/tex]' is the acceleration generated in the roller, then considering the linear motion of the roller, we can write
[tex]&& -F + f = - Ma\\&or,& a = \dfrac{F - f}{M}[/tex]
Considering the rolling motion of the roller, if '[tex]\omega[/tex]' be the angular velocity of the roller, then frictional force '[tex]F_{R}[/tex]' due to rolling is
[tex]F_{R} = I \alpha = (MR^{2})\dfrac{a}{R} = M a R[/tex]
where, '[tex]\alpha[/tex]' is the angular acceleration.
According to the problem, [tex]f = F_{R}[/tex]. Therefore,
[tex]&&a = \dfrac{F - MaR}{M}\\&or,& Ma + MaR = F\\&or,& a = \dfrac{F}{M(1 + R)}[/tex]
