Question

Calculate the following for the situation shown :-
(a)
Speed at D
H
D
pvc = V7gR
a
RC
В.
(b)
Normal reaction at D
(c)
Height Hकैलकुलेट द फॉलोइंग फॉर द सिचुएशन शोन ए स्पीड ऑफ द बी नॉरमल रिएक्शन एंड डीसी हाइट ऑफ एस फॉर द फॉलोइंग गिवेन फिगर ​

Answer 1

Given:

A body is released from a height H , which ends in a circular track of radius R.

To find:

• Speed at point D
• Normal Reaction at point D

Calculation:

We will apply Conservation Of Mechanical Energy principle (at point D w.r.t point C) to get the speed value of block at point D:

$\rm \therefore \: KE1 + PE1 = KE2 + PE2$

$\implies \dfrac{1}{2} m {(v_{C})}^{2} + mgR = \dfrac{1}{2} m {(v_{D})}^{2} + mg(2R)$

$\implies \dfrac{1}{2} m {( \sqrt{7gR} )}^{2} + mgR = \dfrac{1}{2} m {(v_{D})}^{2} + mg(2R)$

$\implies {( \sqrt{7gR} )}^{2} + 2gR = {(v_{D})}^{2} + 4gR$

$\implies 7gR+ 2gR = {(v_{D})}^{2} + 4gR$

$\implies {(v_{D})}^{2} = 5gR$

$\implies v_{D} = \sqrt{5gR}$

So, velocity at point D is √(5gR).

Now according to FBD at point D:

$\therefore \: N + mg = \dfrac{m {(v_{D})}^{2} }{R}$

$\implies \: N + mg = \dfrac{m {( \sqrt{5gR} )}^{2} }{R}$

$\implies \: N + mg = \dfrac{m {(5gR} )}{R}$

$\implies \: N + mg = 5mg$

$\implies \: N = 5mg - mg$

$\implies \: N = 4mg$

So, normal reaction at point D is 4mg.

$\star$ Hope It Helps.

Answer 2

Answer (a):-

As there is no frictional force acting along the track, we can apply conservation of mechanical energy.

For gravitational energy, we can assume any line to be a reference line and say that the Gravitational Potential Energy (GPE) is 0 at that point. And for other points, we can say that the GPE at that point will be [m * g * (displacement between that point and reference line)]

Refer to the first attachment for the reference line which I have considered (that red line).

Then, the GPE at D will be $\sf mgR.$

Now, by conservation of mechanical energy from C to D,

$\sf U_C + K_C = U_D + K_D$

$\longrightarrow \sf 0 + \dfrac{1}{2}m(v_C)^2 = mgR + \dfrac{1}{2}m(v_D)^2$

$\longrightarrow \sf \dfrac{1}{2}m(\sqrt{7gR})^2 = mgR + \dfrac{1}{2}m(v_D)^2$

$\longrightarrow \sf \dfrac{7}{2}mgR - mgR = \dfrac{1}{2}m(v_D)^2$

$\longrightarrow \sf {\cancel{ \dfrac{1}{2}}} \times 5\cancel{m}gR ={\cancel{ \dfrac{1}{2}}}\cancel{m}(v_D)^2$

$\longrightarrow \sf v_D { }^{2} = 5gR$

$\longrightarrow \underline{\underline{\sf{\green{v_D = \sqrt{5gR}}}}}$

Answer (b):-

Refer to the attachment for F.B.D.

$\sf N_D + mg = a_c$

$\longrightarrow \sf N_D + mg = m \times \dfrac{v_D { }^{2}}{R}$

$\longrightarrow \sf N_D + mg = m \times \dfrac{(\sqrt{5gR})^2}{R}$

$\longrightarrow \sf N_D + mg = m \times \dfrac{5g{\cancel{R}}}{{\cancel{R}}}$

$\longrightarrow \sf N_D + mg = 5mg$

$\longrightarrow \sf N_D = 5mg - mg$

$\longrightarrow \underline{\underline{\sf{\green{N_D = 4mg}}}}$

Answer (c):-

As there is no frictional force acting along the track, we can apply conservation of mechanical energy.

Here, the reference line is the bottom surface.

By conservation of mechanical energy from A to C,

$\sf U_A + K_A = U_C + K_C$

$\longrightarrow \sf mgH + 0 = mgR + \dfrac{1}{2}m(v_C)^2$

$\longrightarrow \sf mgH = mgR + \dfrac{1}{2}m(\sqrt{7gR})^2$

$\longrightarrow \sf mgH = mgR + \dfrac{7}{2}mgR$

$\longrightarrow \sf {\cancel{mg}}(H) = {\cancel{mg}}\Big(R + \dfrac{7}{2}R\Big)$

$\longrightarrow \underline{\underline{\sf{\green{ H = \dfrac{9}{2}R }}}}$