Question

A heavy particle slides under gravity down the inside of a smooth vertical tube held in vertical plane. It starts from the highest point with velocity $\sqrt{2ag}$, where $a$ is the radius of the circle. Find the angular position $\theta$ [ as shown in the figure ] at which the vertical acceleration of the particle is maximum. ​

Answer 1

Answer:

The angular position for which the vertical acceleration tends to it's maximum value is ∅ = cos-¹(2/3).

Explanation:

Suppose that the angular position which corresponds to the maximum acceleration is ∅ and that the velocity is v.

Energy at top (Ei) = 1/2m(√2ag)² + mga

Energy at the assumed position (Ef) = 1/2mv² + mga*cos(∅)

Applying the law of conservation of energy at the two positions of the particle. That is, Ei = Ef

1/2m(√2ag)² + mga = 1/2mv² + mga*cos(∅)

mga + mga = 1/2mv² + mga*cos(∅)

2mga = 1/2mv² + mga*cos(∅)

Divide throughout by m,

2ga = 1/2v² + ga*cos(∅)

1/2v² = ga[2 - cos(∅)]

v² = 2ga[2 - cos(∅)]

v²/a = 2g[2 - cos(∅)]

Thus, Centripetal acceleration = v²/a = 2g[2 - cos(∅)] and hence, the centripetal force is,

Fc = m(v²/a) = 2mg[2 - cos(∅)] ---------(I)

Also, the centripetal force, according to the FBD, is N + mgcos(∅). That is, Fc = N + mgcos(∅) ---------(II)

From equations I and II,

2mg[2 - cos(∅)] = N + mgcos(∅)

4mg - 2mgcos(∅) = N + mgcos(∅)

N = 4mg - 3mgcos(∅)

N = mg(4 - 3cos(∅))

According to the FBD, the vertical force acting on the particle is:

Fv = Ncos(∅) + mg = mg[4-3cos(∅)]*cos(∅) + mg

Fv = mg[{4 - 3cos(∅)}*cos(∅) + 1]

m*a(v) = mg[{4 - 3cos(∅)}*cos(∅) + 1]

a(v) = g[4cos(∅) - 3cos²(∅) + 1] ---------(A)

Now, according to the Application of Derivatives (Grade 12 math), the global maxima for any function f(x) occurs at all those points for which f'(x) = 0.

From equation (A) it is clear, that the vertical acceleration a(v) is a function of cos(∅). Thus, it's maximum value (maxima) occurs at all those points for which the derivative of a(v) wrt cos(∅) is zero.

Differentiating a(v) wrt cos(∅), we get

d(av)/dcos(∅) = g[4 - 2*3*cos(∅) + 0]

d(av)/dcos(∅) = g[4 - 6cos(∅)]

For global maxima, d(av)/dcos(∅) = 0

g[4 - 6cos(∅)] = 0

cos(∅) = 4/6 = 2/3

∅ = cos-¹(2/3)