Physics Answer the following in Brief

Given: \(m =100 g =0.1 kg , r =50 cm =0.5 m , g =9.8 m / s ^2, v _{ L }=7 m / s\)
To find:Velocity at the highest point \(v _{ H }\)
\(v _{ H }=\sqrt{\frac{2\left( TE _{ H }\right)}{ m }-4 gr }\)
Calculation:Total energy at highest point,
T.E \(( H )= K \cdot E\) at lowest point \(=\frac{1}{2} mv _{ L }^2\)
\(\therefore T . E _{ H }=\frac{1}{2} \times 0.1 \times 7^2\)
\(=2.45 J\)
From formula,
\(v _{ H }=\sqrt{\frac{2(2.45)}{0.1}-4 \times 9.8 \times 0.5}\)
\(=\sqrt{29.4}=5.422 m / s\)
The velocity at the highest point is \(5.422 m / s\)

• Motion of a car on a banked road:

For the vehicle to go round the curved track at a reasonable speed without skidding, the greater centripetal force is managed for it by raising the outer edge of the track a little above the inner edge. It is called banking of circular tracks.

Consider a vehicle of weight Mg, moving round a curved path of radius r, with a speed v, on a road banked through angleθ.
The vehicle is under the action of the following forces:

• The weight Mg acting vertically downwards
• The reaction R of the ground to the vehicle, acting along the normal to the banked road OA in the upward direction

The vertical component R cos θ of the normal reaction R will balance the weight of the vehicle and the horizontal component R sin θ will provide the necessary centripetal force to the vehicle. Thus,
R cosθ = Mg …(i)
\(R \sin \theta=\frac{M v^2}{r} \ldots \ldots .\). (ii)
On dividing equation (ii) by equation (i), we get
\(\frac{R \sin \theta}{R \cos \theta}=\frac{M v^2 / r}{M} g\)
\(\tan \theta=\frac{v^2}{r g}\)
As the vehicle moves along the circular banked road OA, the force of friction between the road and the tyres of the vehicle, F = μR, acts in the direction AO.
The frictional force can be resolved into two components:

• μ R sinθ in the downward direction
• μ R cosθ in the inward direction

Since there is no motion along the vertical,
R cos θ = Mg + μ R sinθ ……. (iii)
Let vmax be the maximum permissible speed of the vehicle. The centripetal force is now provided by the components R sinθ and μ Mg cosθ, i.e.,
\(R \sin \theta+\mu R \cos \theta=\frac{M v_{\max }^2}{r}\).....(iv)
From equation(iii),we have
Mg = R cosθ (1−μ tanθ)…(v)
Again from equation (iv), we have
\(\frac{M v_{\max }{ }^2}{r}\)
= R cosθ (μ + tanθ) …(vi)
On dividing equation (iv) by (v), we have
\(\frac{v_{\max }^2}{g r}=\frac{\mu+\tan \theta}{1-\tan \theta}\)
\(\Rightarrow v_{\max }=\left(\left(g r \frac{\mu+\tan \theta}{1-\mu \tan \theta}\right)^{1 / 2}\right.\)
Maximum optimum speed depends on:
1) Radius of the curved path,
2) Coefficient of friction
3) angle on inclination
Regards

\(14 m / s ; 1.796 s\)

Lowest point of circular motion:
Total energy = KE
\(=\frac{1}{2} m v^2\)
Substitute \(v =\sqrt{5 rg }\)
Total energy \(=\frac{1}{2} m(\sqrt{5 r g})^2\)
\(=\frac{5 m r g}{2}\)
Highest point of circular motion:
Total enery = PE + KE
= mgh + KE
\(=m g(2 r)+\frac{1}{2} m v^2\)
Substitute \(v =\sqrt{r g}\)
Total energy \(=2 mgr +\frac{1}{2} m(\sqrt{r g})^2\)
\(=2 m g r+\frac{m g r}{2}\)
\(=\frac{5 m r g}{2}\)

Total Rolling kinetic energy = Translational K. E. + Rotational K. E.
\(=\frac{1}{2} M V^2+\frac{1}{2} I \omega^2\)
But
\(\omega=\frac{V}{R}\)
Total Rolling K.E \(=\frac{1}{2} M V^2+\frac{1}{2} I\left(\frac{V^2}{R^2}\right)\)
for solid sphere \(I=\frac{2}{5} M R^2\)
Total Rolling K.E. \(=\frac{1}{2} M V^2+\frac{1}{2} \frac{2 M R^2}{5}\left(\frac{V^2}{R^2}\right)\)
\(=\frac{1}{2} M V^2+\frac{1}{5} M V^2\)
\(=\frac{7}{10} M V^2\) 

The kinetic energy of an object is the energy that it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity.
Consider a body of mass "m" starts moving from rest. After a time interval "t" its velocity becomes V.
If initial velocity of the body is Vi = 0 ,final velocity Vf = V and the displacement of body is "d". Then Derivation for the equation of Kinetic Energy:
The relation connecting the initial velocity (u) and final velocity (v) of an object moving with a uniform acceleration a, and the displacement, S is
v2 - u2 = 2aS
This gives
S = ½a(v2 - u2)
We know F = ma. Thus using above equations, we can write the workdone by the force, F as
W = ma × ½a(v2 - u2)
or
W =m( v 2 - u 2 ) ½
If object is starting from its stationary position, that is, u = 0, then
W = ½mv2
It is clear that the work done is equal to the change in the kinetic energy of an object.
If u = 0, the work done will be W = ½mv2
Thus, the kinetic energy possessed by an object of mass, m and moving with a uniform velocity, v is Ek= ½mv2

I \(=0.625 kg \cdot m ^2 ; \tau=1.25 N \cdot m\)

\(0.08163 m\)

80 r.p.m