91影视

Mass of block = 500 g =0.5 kg
Length of rod = 2 m
Thrust force = 3.5 N.
The nozzle is at an angle of 70^o from the radial line as shown.
The block starts from rest

The centripetal force is calculated by

mv²/r 鈥︹�︹�︹�︹�︹�︹�︹��..(1)

The angular velocity of the mass revolving around a circle is calculated by

${蝇}^2={蝇}_0^2+2伪胃...............................\left(2\right)$

Now break the thrust in tangential and radial components

$$\begin{gathered} F_t=3.5sin(70掳)N \\ {F}_r=3.5cos(70掳)N \end{gathered}$$

Now find angular acceleration of the block

$伪=\frac{F_t}{mr}$

Substitute values

$伪=\frac{(3.5N)\sin (70掳)}{(0.5kg脳2m)}=3.289rad/s^2$

Substitute in equation (2) to get angular velocity after 10 rev

$$\begin{gathered} {蝇}^2=0^2+2(3.289rad/s^2)\left(20蟺rad\right) \\ 蝇=20.33rad/s \end{gathered}$$

Mass of block = 500 g =0.5 kg
Length of rod = 2 m
Thrust force = 3.5 N.
The nozzle is at an angle of 70^o from the radial line as shown.
The block starts from rest.

The centripetal force acting on the block is calculated by substituting values in equation (1) $\left(m{v}^2\right)/r$

substitute v= 蝇r

$\frac{m{v}^2}{r}=\frac{\left(m\right(r蝇{)}^2)}{r}=mr{蝇}^2$

Tension in the rod after 10 rev is

$$\begin{gathered} T=mr{蝇}^2-F_r \\ Substitutevalues \\ T=(0.5kg)\left(2m\right)\left(2\right(3.289rad/s^2)\left(20蟺rad\right)-3.5\cos (70掳) \\ T=412.11N \end{gathered}$$