A mass m is released with horizontal speed v from the top of a smooth and fixed hemispherical bowl of radius r the angle theta with respect to the vertical where it leaves contact the bowl is
Answers
Explanation:
Given in the question :-
The particle is released on the smooth sphere .
For (a)
Hence the Normal force on the surface = N
weight of the particle = mgcosθ
Now equating the conditions
N = mgcos30°
N = √3/2 (mg) .
Hence, Force exerted by the sphere is √3/2(mg)
For (b)
The particle makes the angle θ when it leaves the surface, so N= 0
Hence the velocity v ,
\frac{mv^2}{R} = mgcos \theta
R
mv
2
=mgcosθ
v^2 = Rgcos( \theta + 30^0)v
2
=Rgcos(θ+30
0
) ----(i) .
Now change in Potential energy
= mg{Rcos30^0-Rcos(\theta+30^0)}mgRcos30
0
−Rcos(θ+30
0
) ----(ii)
Change in Kinetic Energy
= \frac{1}{2} mv^{2}
2
1
mv
2
-----(iii)
Now equating equation (ii) & (iii) .
1/2 mv^2 = mg{Rcos30^0-Rcos(\theta+30^0)}1/2mv
2
=mgRcos30
0
−Rcos(θ+30
0
)
v^2 = 2g{Rcos30^0-Rcos(\theta+30^0)}v
2
=2gRcos30
0
−Rcos(θ+30
0
)
Now from here , put the value of v² in equation (i)
gRcos(\theta+30^0)=2g{Rcos30^0-Rcos(\theta+30^0)}gRcos(θ+30
0
)=2gRcos30
0
−Rcos(θ+30
0
)
3cos(\theta+30^0) = \sqrt{3}3cos(θ+30
0
)=
3
cos(θ+30°) = √3/3 = 1/√3
1/√3 ≈cos55°
Now
θ = 55°-30°
θ = 25°
or 0.43 radian
Hence the distance covered by the particle 0.43 radian.
Hope it Helps :-)