Question

A hemisphere of radius 80 mm is cut out from a right circular cylinder of diameter 80mm and height 160 mm as shown in Fig. 6.33. Find the centre of gravity of the bodyfrom the base AB .

Answer 1

Volume of right circular cylinder =
\pi \: r {}^{2} hπr2h 
=22/7×14×14×28 =
17248cm {3}^{}17248cm3 
Volume me of hemisphere =2/3πr×r×r
=2/3×22/7×14×14×14
=
5749.3cm {}^{3}5749.3cm3 
Total volume of solid =17248+5749.3
=
22997.3cm {}^{3}22997.3cm3 

Answer 2

Answer:

A hemisphere of the radius $80mm$ is cut out from a right circular cylinder of diameter $80mm$ and height $160mm$ . The centre of gravity of the body is 67mm from the base $AB$ .

Explanation:

Given:

A right circular cylinder with a diameter of $80mm$ and a height of $160mm$  is cut into a hemisphere with a radius of $80mm$ .

To Solve:

Because the centroid is located on the axis of symmetry and the supplied figure is symmetric along the $y$ axis, we will determine $\bar{y}$ .

The centroid of the specified cone can be determined using the following formulae.

$\bar{y}=\frac{\sum v_i y_i}{\sum v_i}$

where $v_{i}$ is the volume and $y_{i}$ is the angle between the centroid of each individual volume and axis $AB$ .

$V_1=\pi \mathrm{r}^2 \mathrm{~h}=\pi 40^2 160$ (The radius of cylinder $=\frac{80}{2}=40 \mathrm{~mm}$ and height $=160 \mathrm{~mm}$  )

$V_1=804247.72 \mathrm{~mm}^3$ and $y_1=\frac{160}{2}=80 \mathrm{~mm}$

Now the value of $V_2=\frac{2 \pi r^3}{3}=\frac{2 \pi 40^3}{3}=134041.3 \mathrm{~mm}^3$  (The radius of hemisphere is $40mm$ ).

$y_2=160-\frac{3 r}{8}=145 \mathrm{~mm}$

Thus, $\bar{y}=\frac{V_1 y_1-V_2 y_2}{V_1-V_2}=67 \mathrm{~mm}$from the base of $AB$ .

The centre of gravity of the body is   from the base $AB$ .

To know more about "hemisphere"

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