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Losing Seconds – Solve Problem and Learn

Let’s discuss the problem based on losing seconds where we find the depth below the earth’s surface. Try yourself and then read the solution.

The Problem: Losing Seconds

Taking the earth’s radius as (6400 Km) and assuming that the value of (g) inside the earth is proportional to the distance from the earth’s centre, at what depth below the earth’s surface would a pendulum which beats seconds at the earth’s surface lose (5) min in a day?

Discussion:

Let us assume (g) is the acceleration due to gravity at depth (d) below the surface of the earth and (g) is the acceleration due to gravity on the surface.

Let the corresponding time periods be (T_0) and (T).
$$g=g_0(1-\frac{d}{R})$$
where (g) and (g_0) are the acceleration due to gravity at depth respectively, and (R) is the radius of the earth
$$T=T_0\sqrt{\frac{g_0}{g}}=T_0(1-d/R)^{-1/2}$$ $$=T_0(1+\frac{d}{2R})$$
Time registered for the whole body will be proportional to the time period. Thus,
$$\frac{T}{T_0}=\frac{t}{t_0}=1+d/2R$$
$$= \frac{86400}{86400-300}=1+\frac{d}{2R}$$
Substituting
(R=6400Km), we find (d=44.6Km)

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Time Period of a Rolling Cylinder

In this post, we have discussed a problem based on the time period of a rolling cylinder. Try the problem yourself first, then read the solution.

The Problem: Time Period of a Rolling Cylinder

A solid uniform cylinder of radius (r) rolls without sliding along the inside the surface of a hollow cylinder of radius (R), performing small oscillations. Determine time period.

Solution:

Translational kinetic energy + rotational kinetic energy + potential energy= constant

$$\frac{1}{2}mv^2+{\frac{1}{2}I\omega^2+mg(R-r)(1-cos\theta)}=C$$
Now $$I=1/2mr^2$$
$$3/4m(dx/dt)^2+mg(R-r)\theta^2/2=C$$
Differentiating with respect to time,

$$\frac{3}{2}m(\frac{d{^2}x}{dt{^2}})^+mg(R-r)\theta\frac{d\theta}{dt}$$
Now, $$x=(R-r)\theta$$
$$\frac{3}{2} d^2x/dt^2(R-r)d\theta/dt+gxd\theta/dt=0$$
Cancelling (\frac{d\theta}{dt}) throughout

$$\frac{d^2x}{dt^2}+\frac{2}{3}\frac{gx}{R-r}=0$$
this is the equation for SHM, with
$$\omega^2=\frac{2}{3}\frac{g}{R-r}$$
$$T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{3(R-r)}{2g}}$$

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Orbit of Planet (KVPY ’10)

Let’s discuss a problem where we will find the change in the orbit of the planet. Give it a try first, then read the solution.

The Problem:

A planet of mass (m) is moving around a star of mass (M) and radius (R) in a circular orbit of radius (r). The star abruptly shrinks to half its radius without any loss of mass. What change will there in the orbit of the planet?

Discussion:

A planet of mass (m) is moving around a star of mass (M) and radius (R) in a circular orbit of radius (r). The star abruptly shrinks to half its radius without any loss of mass.

As we know, the formula for centripetal force is $$F=\frac{mv^2}{r}$$

Since the radius of the star is decreasing without any change in mass will not have any effect on force exerted by star on planet which is the required centripetal force.

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Angular Velocity of Ice Skater

An ice skater spins at (4\pi )rad/s with her arms extended. If her moment of inertia with arms folded is (80\%) of that with her arms extended, what is her angular velocity when she folded her arms?

Discussion:

Conservation of angular momentum gives $$I_1\omega_1=I_2 \omega_2$$
$$(I_1) (4\pi)=0.8 I_1 \omega_2$$
Hence, (\omega_2=5\pi)

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Angular Momentum

Let’s discuss a problem on angular momentum and prove that the tension in the thread is inversely proportional to the cube of the distance from the hole.

The Problem:

A small mass m tied to a piece of thread moves over a smooth horizontal plane/ The other end of the thread is drawn through a hole with constant velocity. Show that the tension in the thread is inversely proportional to the cube of the distance from the hole.

Discussion:

Angular momentum of the mass is assumed to be constant. The particle velocities (v) and (r) are perpendicular. The angular momentum $$J=mvr=constant$$
Hence,
$$v\propto 1/r$$
The tension T arises from centripetal force $$T=\frac{mv^2}{r}$$ Hence
$$T\propto \frac{1}{r^2}\frac{1}{r}$$ or
$$T\propto 1/r^3$$

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Kinetic Energy of a Load Suspended by Spring

A heavy load is suspended on a light spring with upper half rigidity (2k). The spring is slowly pulled down at the midpoint (a certain work (A) is done thereby) and then released. Determine the maximum kinetic energy (W_k) of the load in subsequent motion.

Solution:

If the middle of the spring is stretched out by a distance (x) while doing work (A), the entire spring is stretched out by (x).
Hence, the potential energy of the spring which is equal to the kinetic energy in the subsequent vibrational motion is $$W_k=kx^2/2$$
When the spring is pulled downwards at the midpoint, only its upper half (whose rigidity is (2k)) is stretched, and the work equal to the potential energy of the extension of the upper part of the spring is $$A=2kx^2/2=kx^2$$. Hence, we may conclude that the maximum kinetic energy of the load in the subsequent motion is $$W_k=A/2$$

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The cosmonauts who landed at the pole found that the force of gravity there is (0.01) of that on the Earth, while the duration of the day on the planet is the same as that on Earth. It turned out that besides that the force of gravity on the equator is zero. Find the radius (R) of the planet.

Discussion:

For a body of mass (m) resting on the equator of a planet of radius (R), which rotates at an angular velocity (\omega), the equation of motion has the form $$m\omega^2R=mg’-N$$ where (N) is the normal reaction of the planet surface, and (g’=0.001g) is the free-fall acceleration on the planet.
The bodies on the equator are assumed to be weightless i.e. (N=0).
We know, (w=2\frac{\pi}{T}), where (T) is the period of revolution of the planet.
Hence we obtain $$R=\frac{T^2}{4\pi^2}g’$$
Substituting the value of (T=8.610^4s) and (g’=0.1m/s^2), we get $$R=1.810^7 Km$$

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Time Periods of Revolution of Two Stars

Let’s discuss a problem where we find out the time periods of revolution of two stars.

The problem: Time Periods of Revolution of Two Stars

The masses of two stars are (m_1) and (m_2) and their separation is (l). Determine the period (T) of their revolution in circular orbits about a common centre.
Since the system is closed, the stars will rotate about their common centre of mass in concentric circles. The equation of motion for the stars will have the form $$m_1\omega_1^2l_1=F$$ and $$m_2\omega_2^2l_2=F……(1)$$
Here (\omega_1) and (\omega_2) are the angular velocities of rotation of the stars, (l_1) and (l_2) are the radii of their orbits, (F) is the force of interaction between the stars, equal to (\frac{Gm_1m_2}{l^2}) where (l) is the seperation between the stars and (G) is the gravitational constant.
By the definition of centre of mass,
$$m_1l_1=m_2l_2$$
$$l_1+l_2=l…… (2)$$
Solving equations 1 and 2 together, we get
$$\omega_1=\omega_2=\sqrt{\frac{G(m_1+m_2)}{l^3}}$$
The required period of revolution of these stars is $$T=2\pi l\sqrt{\frac{l}{G(m_1+m_2)}}$$

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Block on a Conveyor Belt

A conveyor belt having a length (l) and carrying a block of mass (m)
moves at a velocity (v). Determine the distance covered by the block before it stops.

Discussion:

The initial velocity of the block relative to the ground is determined from the conditions $$v_0=at$$ $$l=v_0t-at^2/2$$
Now, acceleration of the block due to friction $$a=\mu g$$
Hence, $$v_0=\sqrt{2\mu gl}$$
The time of the motion of the block along the conveyor belt $$t=\sqrt{\frac{2l}{\mu g}}$$
The distance covered by the block before it stops $$s=l+vt=l+v\sqrt{\frac{2l}{\mu g}}$$

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Moment of Inertia

Four small spheres, each of which you can regard as a point of mass (0.2)Kg are arranged in a square (0.4)m on a side and connected by extremely light rods. Find the moment of inertia of the system about an axis through the centre of the square.
Discussion:

The length of each side of a square is (0.4)m. Thus, the perpendicular bisector is at (0.2)m. The length of the bisector dropped from the point of intersection of the diagonals is (0.2)m. Therefore, the distance of the vertices from the point of intersection of diagonals is $$r=\sqrt{(0.2)^2+(0.2^2)}=0.2828m$$
The moment of inertia $$I=MR^2= 4(0.2)(0.2828)=0.0640kg m^2$$