Question

A spool, with a thread wound on it, of mass $m$ , rests on a rough horizontal surface. Its moment of inertia, relative to its own axis, is equal to, $I = γ m R^{2}$ , where $γ$ is a numerical factor and $R$ is the outside radius of the spool. The radius of the wound thread layer is equal to $r$ . The spool is pulled, without sliding, by the thread, with a constant force $F$ , directed at an angle $α$ to the horizontal. Find:
$(a)$ the projection of the acceleration vector of the spool axis, on the $x$ -axis;
$(b)$ the work performed by the force $F$ , during the first $t second$ , after the beginning of the motion.

Accepted Answer

$(a)$ First, we draw a free body diagram of the spool. The forces acting on the spool are gravitational force $m g$ , force $F$ , friction $γ$ and normal reaction $N$ , at the surface. We consider the positive direction along +ve X-axis. linear acceleration, $ω_{cx}$ and angular acceleration $β_{z}$ , have identical signs. Equation of dynamics gives:

$F \cos (α) - f_{r} = m ω_{cx} . . . (1)$

$f_{r} R - F r = I_{c} β_{z} = γ m R^{2} β_{z} . . . (2)$

To prevent sliding of the spool, $ω_{cx} = β_{z} R . . . (3)$

From the three equations, $ω_{cx} = \frac{F [\cos (α) - (r / R)]}{m (1 + γ)}$

Note: From above expression, we must have $R$

$(b)$ Since work done by the friction is zero, the total work done by $F$ is equal to the gain in kinetic energy

$W_{F} = K E_{f} - K E_{i}$

$W_{F} = K E_{f} = K E_{T} + K E_{R}$

$W_{F} = \frac{1}{2} m v_{c}^{2} + \frac{1}{2} I ω^{2} = \frac{1}{2} m v_{c}^{2} + \frac{1}{2} γ m R^{2} \frac{v_{c}^{2}}{R^{2}} \Rightarrow W_{F} = \frac{1}{2} m (1 + γ) v_{c}^{2}$

Using the kinematical equation, the velocity of the center of mass of the spool, at any time $t$ is,
$v_{c} = 0 + ω_{cx} t$ ,
then

$W_{F} = \frac{1}{2} m (1 + γ) {(ω_{cx} t)}^{2}$

Substituting the value of $ω_{cx}$ , we get

$W_{F} = \frac{F^{2} {(\cos (α) - \frac{r}{R})}^{2} t^{2}}{2 m (1 + γ)}$

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