Question

Please solve q4 fast tomorrow is my online test please help me

Answer 1

Answer:

Given :

Speed at point C = √(7gR)

Solution:

Inorder to find speed at D, we shall apply Conservation of Mechanical Energy concept :

KE1 + PE1 = KE2 + PE2

$= > \frac{1}{2} m {( \sqrt{7gR}) }^{2} + mgR = \frac{1}{2} m {v}^{2} + mg(2R)$

$= > \frac{1}{2}m(7gR) + mgR = \frac{1}{2} m {v}^{2} + mg(2R)$

$= > m {v}^{2} = 5mgR$

$= > v = \sqrt{5gR}$

Velocity at D will be √5gR.

For normal Reaction at D :

$N + mg = \dfrac{m {v}^{2} }{R}$

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

$= > N = 5mg - mg$

$= > N = 4mg$

Normal Reaction at D = 4mg .

For height H :

Again applying Conservation of Mechanical Energy :

KE1 + PE1 = KE2 + PE2

$= > 0 + mgH = \frac{1}{2} m {( \sqrt{7gR}) }^{2} + mgR$

$= > 0 + mgH = \frac{7}{2} mg R + mgR$

$= > H = \dfrac{9}{2} R$

So H = 9/2 R

Answer 2

$\tt{\huge{\pink{Hello!!! }}}$

The above image shows the required figure.

On the above figure we have to apply the conservation of net energy at point C and D

So the Required Equation Becomes :

$\tt{\purple{ \dfrac{1}{2} \times m {V_{C}}^2 + mgr \: = \dfrac{1}{2} \times m {V_{D}}^2 + 2mgr }}$

Solving the above equation we get :

$\tt{\green{V_{D} \: = \sqrt{5gR} }}$

$\tt{\blue{N = \dfrac{m . {V_{D}}^2 } {R} \: - \: mg = 9mg }}$

Again applying Conservation of Mechanical Energy :

KE1 + PE1 = KE2 + PE2

$0 + mgH = \dfrac{1}{2} \sqrt {7gR} + mgr$

Solving Similarly we get :

$H = \dfrac{9}{2} R$