Physics Laws of Motion

$f=\frac{m{v}^2}{R}$            

$\Rightarrow \frac{m{v}^2}{R}=\mu N$

 

$\frac{m{v}^2}{R}=\mu mg$

$\Rightarrow 30=\sqrt[]{\mu *75*10}$                

  $\mu =\frac{30*30}{750}=1.2$              

              Now, R = 48 m (new Radius of curvature)

    $\mu =1.2$

$V_{max}=\sqrt[]{\mu Rg}$

$=\sqrt[]{1.2*48*10}$            

$=24m/sec$               

F. B. D of Beaker

By NLM2

$\begin{array}{l}f=ma=m{\omega }^{2}R\\ \Rightarrow m{\omega }^{2}R\le \mu N\\ R\le \mu g/{\omega }^2\end{array}$                              

m₁ = 5kg

m₂ = 3kg

As m₁ is in rest

T = m_gg = 30

N = m₁ g cos

T = $m_{1}gsin\theta \Rightarrow 50sin\theta =30$

sin $=\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\Rightarrow \theta =37°$

$\therefore N=50cos\theta =50cos37°=50*\frac{4}{5}=40N$

The two conditions for equilibrium in physics are

1. Translational Equilibrium: The net force acting on the object must be zero, which means all the horizontal and vertical forces cancel each other out.
2. Rotational Equilibrium: The net torque must also be zero, indicating that the clockwise and counter-clockwise torques are balanced. 

To reach equilibrium, all the forces and torques acting on an object need to balance each other out. In short, every action force must be met with an equal and opposite reaction. Static equilibrium happens when the object is at rest. Dynamic equilibrium is when it's moving at a constant velocity. Usually, we draw a free-body diagram to visualise these forces. Further, to prove that the object has reached equilibrium state, we use Newton's laws to prove that both the net force and net torque equal zero. 

Equilibrium in physics tells us about a situation where an object's motion or resting situation does not change. This can mean the object is either at rest (static) or moving at a steady speed (dynamic). This state occurs when there's no net external force or torque acting on the object. This thought follows Newton's First Law, which we know that an object will maintain its state of motion unless a force acts upon it. 

Assume both the block is moving with common acceleration 'a' then

$a=\frac{F}{m+M}.......(i)$

N = mg

N = 2g

$f=ma=\frac{mF}{m+M}$

For no slipping

$\Rightarrow \frac{2F}{10}\le g$

$F\le 5g$

$\begin{array}{l}F\le 5*9.8\\ F\le 49.0N\end{array}$

From the physics perspective, in uniform circular motion, speed and angular velocity are constant. That makes the angular acceleration and tangential acceleration are zero. Non-uniform motion, on the other hand, has varying speed and angular velocity, where angular acceleration is non-zero. 

$\frac{\mathrm{T}}{\sqrt[]{2}}=100\mathrm{}\mathrm{}\frac{\mathrm{T}}{\sqrt[]{2}}=\mathrm{F}$

$F=100N$

The period T is the time taken to complete one full revolution (2? radians). In uniform circular motion, it relates to angular velocity by? =2? /T and to tangential speed by v=? r=2? r/T .