• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon

Suggested languages for you:

Americas

Europe

Q99P

Expert-verified
Found in: Page 91

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# In Fig. 4-54, a lump of wet putty moves in uniform circular motion as it rides at a radius of 20.0cm on the rim of a wheel rotating counterclockwise with a period of 5.00 ms .The lump then happens to fly off the rim at the 5 o’clock position (as if on a clock face). It leaves the rim at a height of h=1.20m from the floor and at a distance d=2.50 m from a wall. At what height on the wall does the lump hit?

The height of the wall from the floor at lump will hit is 2.64 m.

See the step by step solution

## Step 1: The given data

$1\right)\mathrm{Radius}\mathrm{of}\mathrm{wheel},\mathrm{r}=20.0\mathrm{cm}\mathrm{or}0.2\mathrm{m}\phantom{\rule{0ex}{0ex}}2\right)\mathrm{Period}\mathrm{of}\mathrm{lump},\mathrm{T}=5.00\mathrm{ms}\mathrm{or}5×{10}^{-3}\mathrm{s}\phantom{\rule{0ex}{0ex}}3\right)\mathrm{Height}\mathrm{of}\mathrm{lump}\mathrm{while}\mathrm{leaving}\mathrm{the}\mathrm{rim},\mathrm{h}=1.20\mathrm{m}\phantom{\rule{0ex}{0ex}}4\right)\mathrm{Distance}\mathrm{of}\mathrm{wall}\mathrm{from}\mathrm{the}\mathrm{wheel},\mathrm{d}=2.50\mathrm{m}$

## Step 2: Understanding the concept of the projectile motion

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, $\mathrm{y}={\mathrm{tan\theta }}_{0}\mathrm{d}-\frac{{\mathrm{gd}}^{2}}{2{\left({\mathrm{V}}_{0}{\mathrm{cos\theta }}_{0}\right)}^{2}}$…(i)

The velocity of a body in circular motion, $V=\frac{2\pi r}{T}$ …(ii)

## Step 3: Calculation of the height of the wall

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:

$\mathrm{V}=\frac{2\mathrm{\pi }\left(0.2\mathrm{m}\right)}{\left(5×{10}^{-3}\mathrm{s}\right)}\phantom{\rule{0ex}{0ex}}=251.32\mathrm{m}/\mathrm{s}$

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:

${\mathrm{\theta }}_{0}=\frac{1\mathrm{hr}×360°}{12\mathrm{hr}}\phantom{\rule{0ex}{0ex}}=30°$

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

$\mathrm{y}=\mathrm{tan}\left(30°\right)\left(2.5\mathrm{m}\right)-\frac{\left(9.8\mathrm{m}/{\mathrm{s}}^{2}\right){\left(2.5\mathrm{m}\right)}^{2}}{2\left(\left(251.32\mathrm{m}/\mathrm{s}\right)\mathrm{cos}\left(30°\right)}\phantom{\rule{0ex}{0ex}}\approx 1.44\mathrm{m}$

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