91Ó°ÊÓ

$$\begin{gathered} 1)\mathrm{Radius}\mathrm{of}\mathrm{wheel},\mathrm{r}=20.0\mathrm{cm}\mathrm{or}0.2\mathrm{m} \\ 2)\mathrm{Period}\mathrm{of}\mathrm{lump},\mathrm{T}=5.00\mathrm{ms}\mathrm{or}5×{10}^{-3}\mathrm{s} \\ 3)\mathrm{Height}\mathrm{of}\mathrm{lump}\mathrm{while}\mathrm{leaving}\mathrm{the}\mathrm{rim},\mathrm{h}=1.20\mathrm{m} \\ 4)\mathrm{Distance}\mathrm{of}\mathrm{wall}\mathrm{from}\mathrm{the}\mathrm{wheel},\mathrm{d}=2.50\mathrm{m} \end{gathered}$$

The motion of an object hurled or projected into the air is called projectile motion since it is only affected by gravity's acceleration. Here, we have to consider the two types of motion. When the lump is moving along the rim, it will have a uniform circular motion. While, when it flies off the rim, it will undergo projectile motion.

Formulae:

The trajectory of a particle in a projectile motion, $\mathrm{y}={\mathrm{³Ù²¹²Ôθ}}_0\mathrm{d}-\frac{{\mathrm{gd}}^2}{2{\left({\mathrm{V}}_0{\mathrm{³¦´Ç²õθ}}_0\right)}^2}$…(¾±)

The velocity of a body in circular motion, $V=\frac{2\mathrm{Ï€}r}{T}$ …(¾±¾±)

Using r=0.2 m and$T=5×{10}^{-3}s$ in equation (ii), we get the speed of a lump while revolving with rim as follows:

$$\begin{gathered} \mathrm{V}=\frac{2\mathrm{π}\left(0.2\mathrm{m}\right)}{\left(5×{10}^{-3}\mathrm{s}\right)} \\ =251.32\mathrm{m}/\mathrm{s} \end{gathered}$$

The lump leaves the rim at 5 o’clock it means it is 1 hour away from bottom position, so the angle of the lump is given as:

$$\begin{gathered} {\mathrm{θ}}_0=\frac{1\mathrm{hr}×360°}{12\mathrm{hr}} \\ =30° \end{gathered}$$

Now, using these data in equation (i), we get the height of projection as follows:

$$\begin{gathered} \mathrm{y}=\tan \left(30°\right)\left(2.5\mathrm{m}\right)-\frac{\left(9.8\mathrm{m}/{\mathrm{s}}^2\right){\left(2.5\mathrm{m}\right)}^2}{2\left((251.32\mathrm{m}/\mathrm{s}\right)\cos \left(30°\right)} \\ ≈1.44\mathrm{m} \end{gathered}$$

So, the total height from the floor will become: 1.20 m+1.44 m=2.64 m

Hence, the value of the height is 2.64 m