Question

A large globe, with a radius of about 5 ${\rm m}$ , was built in Italy between 1982 and 1987. Imaginethat such a globe has a radius and a frictionless surface. A small block of mass slides starts from rest at the very top of theglobe and slides along the surface of the globe. The block leavesthe surface of the globe when it reaches a height $h_crit$ above the ground. The geometry of the situation is shownin the figure for an arbitrary height .

$MWE_we_9.jpg$

part A

Consider what happens at the moment when theblock leaves the surface of the globe. Which of the followingstatements are correct?

1. The net acceleration of the block is directed straightdown.
2. The component of the force of gravity toward the center of theglobe is equal to the magnitude of the normal force.
3. The force of gravity is the only force acting on theblock.

part B

Which of the following statements is alsotrue at the moment when the block leaves the surface of theglobe?

part C

Using Newton's 2nd law, find $v_crit$ , the speed of the block at the critical moment when theblock leaves the surface of the globe.

Assume that the height at which the block leaves the surface ofthe globe is $h_crit$ .

Express the speed in terms of, $h_crit$ , and ,the magnitude of the accleration due to gravity. Do notuse in your answer.

part D

Use the law of conservation of energy to find $v_crit$ . This will give you a difference expression for $v_crit$ than you found in the previous part.

Express $v_crit$ in terms of , $h_crit$ , and

part E

Find $h_crit$ , the height from the ground at which the block leaves thesurface of the globe.

Express $h_crit$ in terms of .

Answer 1

Concepts and reason

The concepts used in this problem are Newton’s second law of motion, centripetal acceleration and law of conservation of energy. Calculate the speed using the Newton’s second law of motion. Calculate the speed using the law of conservation of energy.

Fundamentals

Newton’s Second Law of motion:

According to Newton’s second law of motion, the force on an object is equal to the product of mass and acceleration. It depends on mass and acceleration.

The expression of net force is:

$F = ma$

Here, $m$ is the mass and $a$ is the acceleration.

Law of conservation of energy:

According to this law, “the total energy of a system remains conserved.” The kinetic and potential decreases or increases but the total energy is always constant.

Centripetal acceleration:

The linear acceleration is the rate of change of linear velocity but the rate of change of tangential velocity is the centripetal acceleration. The centripetal makes an object to move in a curved path.

The expression for centripetal acceleration is:

${a_{\rm{c}}} = \frac{{{v^2}}}{R}$

Here, ${a_{\rm{c}}}$ is the centripetal acceleration, $v$ is the velocity and $R$ is the radial distance.

Kinetic energy:

When an object is moving with some speed, then it has some kinetic energy. The kinetic energy is:

$K = \frac{1}{2}m{v^2}$

Here, $m$ is the mass of object and $v$ is the speed of object.

Gravitational potential energy:

When an object is at some height, then it said to possess some potential energy. The gravitational potential energy is:

$P = mgh$

Here, $m$ is the mass, $g$ is the acceleration due to gravity and $h$ is the height of the object.

Total energy:

The total energy is the sum of kinetic energy and potential energy of the system.

$E = P + K$

Here, $P$ is the potential energy and $K$ is the kinetic energy.

(A)

The force diagram is:

The mass moves downward, so the net acceleration of the block is straight downward.

As shown in diagram, only the force of gravity acts on the mass.

(B)

As shown in above diagram, the normal force on the mass becomes zero when the mass leaves the surface.

(C)

The net force acting on the mass when the mass leaves the surface of the globe is:

${F_{{\rm{net}}}} = mg\sin \theta$ (Because the normal force is zero at the critical point)

According to Newton’s second law of motion, the net force is:

${F_{{\rm{net}}}} = ma$

Substitute $\frac{{{v_{{\rm{crit}}}}^2}}{R}$ for $a$ .

${F_{{\rm{net}}}} = m\frac{{{v_{{\rm{crit}}}}^2}}{R}$

Substitute $mg\sin \theta$ for ${F_{{\rm{net}}}}$ .

$mg\sin \theta = m\frac{{{v_{{\rm{crit}}}}^2}}{R}$

$\sin \theta = \frac{{{v_{{\rm{crit}}}}^2}}{{Rg}}$ ……(1)

From the figure:

$R + R\sin \theta = {h_{{\rm{crit}}}}$

$\sin \theta = \frac{{{h_{{\rm{crit}}}} - R}}{R}$ ……(2)

From equation (1) and equation (2):

$\begin{array}{c}\\\frac{{{v_{{\rm{crit}}}}^2}}{{gR}} = \frac{{{h_{{\rm{crit}}}} - R}}{R}\\\\{v_{{\rm{crit}}}} = \sqrt {g\left( {{h_{{\rm{crit}}}} - R} \right)} \\\end{array}$

(D)

Total energy of the mass at the initial position:

${E_{\rm{i}}} = {P_{\rm{i}}} + {K_{\rm{i}}}$

Substitute $0$ for ${K_{\rm{i}}}$ and $mg\left( {2R} \right)$ for ${P_{\rm{i}}}$ .

$\begin{array}{c}\\{E_{\rm{i}}} = mg\left( {2R} \right) + 0\\\\ = 2mgR\\\end{array}$

Total energy at the critical point:

${E_{\rm{f}}} = {P_{\rm{f}}} + {K_{\rm{f}}}$

Substitute $\frac{1}{2}m{v_{{\rm{crit}}}}^2$ for ${K_{\rm{f}}}$ and $mg{h_{{\rm{crit}}}}$ for ${P_{\rm{f}}}$ .

${E_{\rm{f}}} = mg{h_{{\rm{crit}}}} + \frac{1}{2}m{v_{{\rm{crit}}}}^2$

According to the law of conservation of energy:

${E_{\rm{i}}} = {E_{\rm{f}}}$

$\begin{array}{c}\\2mgR = mg{h_{{\rm{crit}}}} + \frac{1}{2}m{v_{{\rm{crit}}}}^2\\\\2gR = g{h_{{\rm{crit}}}} + \frac{1}{2}{v_{{\rm{crit}}}}^2\\\\{v_{{\rm{crit}}}}^2 = 2g\left( {2R - {h_{{\rm{crit}}}}} \right)\\\\{v_{{\rm{crit}}}} = \sqrt {2g\left( {2R - {h_{{\rm{crit}}}}} \right)} \\\end{array}$

(E)

The expressions of critical speed are:

${v_{{\rm{crit}}}} = \sqrt {g\left( {{h_{{\rm{crit}}}} - R} \right)}$

${v_{{\rm{crit}}}} = \sqrt {2g\left( {2R - {h_{{\rm{crit}}}}} \right)}$

Equating above equations:

$\begin{array}{c}\\\sqrt {g\left( {{h_{{\rm{crit}}}} - R} \right)} = \sqrt {2g\left( {2R - {h_{{\rm{crit}}}}} \right)} \\\\g\left( {{h_{{\rm{crit}}}} - R} \right) = 2g\left( {2R - {h_{{\rm{crit}}}}} \right)\\\\{h_{{\rm{crit}}}} - R = 4R - 2{h_{{\rm{crit}}}}\\\\{h_{{\rm{crit}}}} = \frac{5}{3}R\\\end{array}$

Ans: Part A

The statements ‘1’ and ‘3’ are correct.

Part B

The statement ‘b’ is true.

Part C

The critical speed of the mass is $\sqrt {g\left( {{h_{{\rm{crit}}}} - R} \right)}$ .

Part D

The critical speed of the mass is $\sqrt {2g\left( {2R - {h_{{\rm{crit}}}}} \right)}$ .

Part E

The expression of critical height is $\frac{5}{3}R$ .