Question

1:- The situation shown in a figure a ball of mass root 2 kg connected to the vertical revolves rod by 2light string of length root2 m each making an angle of 45 degree with the rod both the rod and ball r revolves with angular velocity of roots rad/s the tension in upper string is ________ newtons (nearest integer) ​
ans:-
20 newtons

Answer 1

Answer:

20 Newton's

this is my answar

Answer 2

The tension in upper string is 20N.

Given,

mass of ball(m) = $\sqrt{2}$ kg

length of string(l) = $\sqrt{2}$ m

angle between the rod and the string($\theta$) = $45^\circ$

angular velocity($\omega$) = $\sqrt{10}$ rad/s

Since, the ball is rotating in circular path with constant angular speed

⇒ angular acceleration($\alpha$) = 0 $m/s^{2}$

and, tangential acceleration($a_t$) = 0 $m/s^{2}$

Forces acting on the ball are as follows,

Tension due to first string = $T_1$

Tension due to second string = $T_2$

Weight of ball = mg

As there is no motion in vertical direction

$\sum F_y=0$

⇒ $T_1 \sin 45^\circ - T_2 \sin 45^\circ -mg=0$

⇒ $T_1-T_2=\frac{\sqrt{2}g}{\sin 45^\circ}$

⇒ $T_1-T_2=2g$  

if $g=10$ $m/s^{2}$

⇒ $T_1-T_2=20$   .................................................................................(1)

Centipedal force is provided by the ball as the net force in $x$- direction,

$F_c=(T_1+T_2)\cos 45^\circ$

⇒ $\frac{(T_1+T_2)}{\sqrt{2} } = m\omega^{2} r$  .................................................................................(2)

where,

$r=l\sin 45^\circ$

$r=\sqrt{2} \times\frac{1}{\sqrt{2} }$

$r=1$

Now, putting the value of r in equation (2) we get,

$\frac{T_1+T_2}{\sqrt{2} }=\sqrt{2} \times \sqrt{10} ^{2} \times 1$

⇒ $T_1+T_2= 20$     ...................................................................................(3)

Equation (1) + Equation (3),

$2T_1=40$

∴ $T_1 =20$ N

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