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Q99P

Expert-verifiedFound in: Page 91

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.

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

**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)

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

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

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:

${\mathrm{\theta}}_{0}=\frac{1\mathrm{hr}\times 360\xb0}{12\mathrm{hr}}\phantom{\rule{0ex}{0ex}}=30\xb0$

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

$\mathrm{y}=\mathrm{tan}\left(30\xb0\right)\left(2.5\mathrm{m}\right)-\frac{\left(9.8\mathrm{m}/{\mathrm{s}}^{2}\right){\left(2.5\mathrm{m}\right)}^{2}}{2\left((251.32\mathrm{m}/\mathrm{s}\right)\mathrm{cos}\left(30\xb0\right)}\phantom{\rule{0ex}{0ex}}\approx 1.44\mathrm{m}$

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

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