91Ó°ÊÓ

We have the given data:

Mass of the disk (m) = 40.0 kg.

The radius of the disk (r) =0.200 m.

Force (F)=30.0 N.

The difference between the initial and final angle $\mathrm{θ}-{\mathrm{θ}}_0=0.200$revolutions.

The magnitude of the tangential velocity is given by the formula,

$\mathrm{Ï\dots }=R\mathrm{Ó\neg }⋯⋯\left(1\right)$

In order to find the magnitude of the tangential velocity, first we need to find the angular speed$\mathrm{Ó\neg }$ of the disk after 0.200 revolutions.

The initial angular velocity is 0.

The final angular velocity is given by,

${\mathrm{Ó\neg }}^2=2\mathrm{Î\pm }(\mathrm{θ}-{\mathrm{θ}}_0)⋯⋯\left(2\right)$,

where,$\mathrm{Î\pm }$ is angular acceleration,

$\mathrm{Î\pm }=\frac{\mathrm{Ï„}}{I}⋯⋯\left(3\right)$.

Now, the torque exerted by the force is,

$\mathrm{τ}=RF⋯⋯\left(4\right)$,

and the moment of inertia of the disk about the axis through its center is,

$I=\frac{1}{2}m{R}^2⋯⋯\left(5\right)$

Using all these equations in , we get,

$$\begin{gathered} {\mathrm{Ó\neg }}^2=2\frac{RF}{{\displaystyle \frac{1}{2}}m{R}^2}\left(\mathrm{θ}-{\mathrm{θ}}_0\right) \\ ⇒\mathrm{Ó\neg }=\sqrt{\frac{4F\left(\mathrm{θ}-{\mathrm{θ}}_0\right)}{mR}} \end{gathered}$$

Then equation (1) becomes,

$\mathrm{ν}=R\sqrt{\frac{4F\left(\mathrm{θ}-{\mathrm{θ}}_0\right)}{mR}}=\sqrt{\frac{4FR\left(\mathrm{θ}-{\mathrm{θ}}_0\right)}{m}}$.

Therefore, substituting the values, we get,

$$\begin{gathered} \mathrm{ν}=\sqrt{\frac{4FR\left(\mathrm{θ}-{\mathrm{θ}}_0\right)}{m}} \\ =\sqrt{\frac{4\left(30.0\right)\left(0.200\right)\left(0.200\right).2\mathrm{π}}{40.0}} \\ =0.868 \\ ∴\mathrm{ν}=0.868m/s \end{gathered}$$

where, we have used 1 rev=$2\mathrm{Ï€}$ rad.

Hence, the magnitude of the tangential velocity is,$\mathrm{ν}=0.868$ m/s.

The magnitude of the resultant acceleration is given by,

$a=\sqrt{{a^2}_{\tan }+a_{rad}^2}.........\left(6\right)$,

where,$a_{\tan }$ and$a_{rad}$ are tangential and radial acceleration respectively which are given by,

$a_{\tan }$

$a_{rad}=R{\mathrm{Ó\neg }}^2$

Since we have

$\mathrm{Î\pm }=\frac{RF}{{\displaystyle \frac{1}{2}}m{R}^2}=\frac{2F}{mR}$and ${\mathrm{Ó\neg }}^2=2\frac{RF}{{\displaystyle \frac{1}{2}}m{R}^2}\left(\mathrm{θ}-{\mathrm{θ}}_0\right)=\frac{4F}{mR}\left(\mathrm{θ}-{\mathrm{θ}}_0\right)$, on substituting in above two equations, we get,

$$\begin{gathered} a_{\tan }=R.\frac{2F}{mR}=\frac{2F}{m}, \\ {a}_{rad}=R.\frac{4F}{mR}\left(\mathrm{θ}-{\mathrm{θ}}_0\right)=\frac{4F}{m}\left(\mathrm{θ}-{\mathrm{θ}}_0\right) \end{gathered}$$

Substituting these values in , we get,

$$\begin{gathered} a=\sqrt{{\left(\frac{2F}{m}\right)}^2+{\left(\frac{4F}{m}\left(\mathrm{θ}-{\mathrm{θ}}_0\right)\right)}^2} \\ =\frac{4F}{m}\sqrt{1+2\left(\mathrm{θ}-{\mathrm{θ}}_0\right)} \\ =\frac{1\left(30.0\right)}{40.0}\sqrt{1+4{\left(0.200.2\mathrm{π}\right)}^2} \\ =4.06 \\ ∴a=4.06m/s^2 \end{gathered}$$

m/s².

Hence, the magnitude of the resultant acceleration is, a=4.06 m/s².