Question

A sphere is attached to an inextensible string of length 0.5 m whose upper end is fixed to ceiling. The sphere has mass of 0.2 kg and is made to describe horizontal circle of radius 0.3 m. The speed of sphere is​

Answer 1

Answer:

m=200g

l=130cm

130cm=1.3 metre

So the time period of a pendulum doesn't depend on the mass of the Bob. It is equal to :

g

2π

l

(2pi×rootl/rootg) where l is length and g is gravity. Which gives us:

t2=4π2×1.3/9.8 (squaring both sides)

π2 and 9.8 are approx the same so we can cut it out, giving us:

4×1.3=520s

Therefore t2=5.2

t=2.29s

As per the second part of the question, the force on the string is simple mg = 200×10=2000 gram m/s^2

Or 200gf

Or 0.2kg×10=2N

Answer 2

The speed of the sphere is 1.5 m/s

Explanation:

Given that

$l=0.5$ m

$m=0.2$ kg

$r=0.3$ m

As shown in the attached figure, if the tension in the string is T then the component $T\cos\theta$ will provide the necessary centripetal force

Thus,

$T\cos\theta=m\frac{v^2}{r}$

Where v is the velocity

Also, the vertical component will be balanced by the weight of the sphere

Thus,

$T\sin\theta=mg$

From the above two equations

$\tan\theta=\frac{r}{v^2}$

or, $\frac{\sqrt{l^2-r^2}}{r}=\frac{rg}{v^2}$

$\implies v^2\sqrt{l^2-r^2}=r^2g$

$\implies v^2=\frac{r^2g}{\sqrt{l^2-r^2}}$

$\implies v^2=\frac{0.3^2\times 10}{\sqrt{0.5^2-0.3^2}}$

$\implies v^2=\frac{0.09\times 10}{\sqrt{0.25-0.09}}$

$\implies v^2=\frac{0.9}{0.4}$

$\implies v^2=\frac{9}{4}$

$\implies v=\frac{3}{2}$ m/s

$\implies v=1.5$ m/s

Hope this answer is helpful.

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