Gravitation

Since the moon is in uniform circular motion with earth as the center. Therefore, when the gravitational interaction between the moon and the earth disappears the centripetal force due to which moon is revolving around the earth will become zero and it moves tangentially to the position it was present. Actually, gravitational force doesn’t disappear, so these worries are useless.

The required centripetal force is provided by the normal reaction of the ground.

Now, according to the question;

Radius of circular path (r) = 10 m.

Mass of the car (m) = 1000 Kg.

Speed of the car (v) = 10 m/s.

The figure below describes the situation.

The required centripetal force is F =

⇒ F =

⇒ F =

The centripetal force required for the car is 10⁴ N .

Given;

The radius of hemisphere = R.

The velocity of washer = V.

The centripetal force act on the washer is

F = ………………… (1)

Let the mass of washer is “m”

And acceleration by gravity is “g”

The force by hemisphere on the washer is

F = mg …………. (2)

By Newton’s law of motion every action has an equal reaction but in opposite direction.

∴ Centripetal force on washer = Force by hemisphere on the washer.

From (1) and (2).

= mg;

⇒ v² = gR

⇒ v =

Hence minimum horizontal velocity should be imparted to the washer to detach it from the hemisphere at the initial point of motion is

Carrying a pole helps the walker increase their rotational inertia (a measure of an object’s opposition/resistance to change in its direction of rotation), which aids in maintaining stability while walking over the narrow rope.

The pole also adds more weight below the center of gravity of the walker, which is another bonus for maintaining balance. By adjusting the place of the pole, one can balance easily their body and can walk on the rope. Finally, the line of the weight of total system must pass through a vertical line drawn to the rope.

The Figure below depicts the rope walker carrying the pole on tight rope.

In both cases, you have the same amount of force in the vertical direction, but when we have a single bucket in hand there is a lot of force in the horizontal direction.
With one bucket you are constantly leaning in the other direction so that the downward force is exerting directly on your center of gravity. Holding two buckets of water in each hand does not change the center of gravity of the person.But if the person holds a bucket with the same quantity of water only in one hand, then the center of gravity moves towards the bucket (mass). Change in center of gravity causes instability of body as the gravitational force acts on it. So we feel hard to hold the bucket.

Since Apple dropped from a tree.

∴ Initial speed (u) = 0 m/s

Given in question;

Time (t) = 1.5 s

Acceleration due to gravity (g) = 10 m/s²

Now;

By Newton’s equation of motion

V = u + at.

Here a = g = 10 m/s^2.

T = 1.5 s;

U = 0 m/s

⇒ Final speed (v) = u + gt.

⇒ v = 0 + 10 x 1.5.

⇒ v = 15 m/s

Hence the speed of an apple dropped from a tree after 1.5 second is 15 m/s

By Newton’s equation of motion

S = ut +

Here a = g = 10 m/s^2.

T = 1.5 s;

U = 0 m/s

Distance (S) = ut +

⇒ s = 0 x 1.5 +

⇒ s = 0 +

⇒ s = .25 m.

Hence distance covered by an apple is 11.25 m.

A body is projected vertically up.

∴ According to the question;

The initial speed (u) = 40 m/s

Acceleration due to gravity (g) = 10 m/s² .

From Newton’s third equation of motion;

V² = u² + 2as;

In our question;

V = 0 (Velocity at maximum height is zero);

U = 40 m/s.

A = -g = -10 m/s². (When object will be going up the acceleration due to gravity will be acting downwards to make an object to fall. Hence by sign convention direction of motion and acceleration is opposite therefore a is negative)

S = Maximum Height (H).

Putting the values in the equation we get.

0² = 40² – 2 x 10 x H.

⇒ 0 = 1600 – 20H.

⇒ 1600 = 20H.

⇒ H = .

Maximum height reached by the body is 80m.

Total time (T ) =

⇒ T =

⇒ T =

The entire time of motion is 8 s

Time is taken to reach maximum height =

=

Thus it takes 4 sec to reach the maximum height and returns with 0 initial speed.

∴ speed after 1 sec in return = speed after 5 seconds of throwing the ball.

⇒ speed (v) = u + at;

∵ u = 0 and t = 1sec; a = 10 m/s² .

⇒ v = 0 + 10 x 1

⇒ v = 10 m\s.

Speed after 5 sec of throwing of the ball is 10 m/s.

Since the boy is throwing two balls per second

⇒ He is throwing one ball in half a second.

⇒ Ball takes half second to reach the maximum height.

WE know that time taken to reach maximum height =

⇒

⇒

⇒

⇒

Hence initial velocity of ball is 5 m\s.

Also the maximum height (H) =

⇒ H =

⇒ H =

⇒ H = 1.25 m

Maximum height reached by the ball is 1.25 m

When the man is standing against the wall such that his right shoulder and right leg are along his height. Then the center of

Gravity is nearly at the belly point. In this situation, he can’t raise his left leg. Because if he raises the left leg, the center of gravity moves. Then he must move his body parts to maintain the center of gravity. Therefore as he raises his left leg he has to move his other body parts. The figure below depicts this.

.

Since ball takes 0.2 s to cross the last 6m before hitting the ground.

∴ Let velocity of the ball after time t₁ = u;

The figure below depicts the situation of the question.

By Newton’s second equation of motion;

Distance (S) = ut +

Here S = 6m; t = 0.2s; g = 10m/s²;

⇒ 6 = 0.2u +

⇒ 6 = 0.2u +

⇒ 6 = 0.2u + 0.2

⇒ 0.2u = 6 – 0.2

⇒ 0.2u = 5.8.

⇒ u = 29 m/s.

Hence final speed (v) of the ball after it traveled ‘X’ distance is 29m/s.

By Newton’s first equation of motion

⇒ v = u + at;

Here v = 29 m/s (final velocity).

u = 0 (Initial velocity is zero);

A = g = 10 m/s².

T = t₁ (refer figure above);

Putting the values we get

29 = 0 + 10 t₁;

⇒ t₁ =

Now finding the distance ‘X’ by newton’s second equation of motion

Distance (S) = ut +

Here S = X (refer figure); t = t₁ = 2.9s; g = 10m/s²; u = 0 (initial velocity);

Putting the values we get;

X = 0(2.9) +

⇒ X = 0 +

⇒ X = 5 x 8.41

⇒ X = 42.05 m

Now;

Total height from which ball was thrown = X + Y = 42.05 + 6

⇒ Total height = 48.05 m;

According to question it is given;

Speed of balloon = 5m/s (upward)

⇒ Velocity of ball (u) = -5 m/s (because ball will fall downwards)

Height (S) = 60 m

Time to reach ground (t) = ?

By newton’s second equation of motion

Distance (S) = ut +

⇒ 60 = -5t +

⇒ 60 = -5t + 5t².

⇒ = -t + t².

⇒ 12 = -t + t².

⇒ t² – t – 12 = 0;

⇒ (t – 4) (t – 3) = 0;

⇒ t = 4, -3;

But time can’t be negative;

∴ t = 4 s

Hence Time taken by the ball to reach the ground is 4 s

Since the ball is projected vertically up.

∴ The initial speed (u) = 50 m/s

Acceleration due to gravity (g) = 10 m/s².

Maximum Height (H) =

⇒ H =

⇒ H = 125 m.

Maximum Height is 125 m.

Time to reach maximum height (t₁) =

⇒ t₁ =

Time to reach maximum height is 5 sec.

Speed at maximum height (V) = 0 m/s.

According to the question given that;

Mass of first car = m₁.

The radius of first car = r₁.

Mass of second car = m₂.

The radius of second car = r₂.

Now let;

The speed of first car = V₁.

The speed of second car = V₂.

Time is taken for the first car to complete = T₁ =

Time is taken for the second car to complete = T₂ = _.

According to question;

T₁ = T₂.

⇒ _.

⇒

⇒ .

Hence the ratio of speeds is .

Centripetal acceleration of First car a₁ =

Centripetal acceleration of Second car a₂ =

⇒

⇒

⇒

⇒

Hence the ratio of centripetal acceleration is

The Gravitational force of attraction between two bodies of masses m₁ and m₂separated by a distance of R is given by;

G = Gravitational constant;

The figure below illustrates the question well;

According to our question;

m₁ = ;

m₂ = ;

R = 10 cm = = 0.1 m (1 m = 100 cm).

F =

⇒ F =

⇒ F =

⇒ F =

Hence force between two spherical balls is 6.67 x 10^-7 N

According to the question given data is;

Mass of the moon (M) = 7.4 x 10²² Kg;

Radius of moon (R) = 1740 km = 1740 x 10³ m (1km = 1000m).

Now;

Acceleration due to gravity (g) =

⇒ g =

⇒ g =

⇒ g =

⇒ g =

⇒ g =

⇒ g = 1.63 m\s².

Free fall Acceleration at the surface of the moon is 1.63 m\s²

Now;

Acceleration due to gravity on earth ag_e = 9.8 m\s² .

Acceleration due to gravity on moon g_m = 1.63 m\s² .

⇒

g_e = 6 g_m.

Acceleration due to gravity on the moon is one-sixth of Acceleration due to gravity on earth.

No, we cannot find any two particles which do not exert a gravitational force on each other. Because according to Newton's universal law of gravitation everybody can exert a gravitational force on another body. Hence there are no two bodies which do not exert a gravitational force on each other. If there would have been any such body than it would have violated the Newton's universal law of gravitation.

According to Newton’s law ‘every action has an equal reaction but in opposite direction’. Therefore the apple applies gravitational force on earth which is equal to the force applied

By the earth apple. But the acceleration of earth due to the force of apple is very small. It is negligible due to the large mass of the earth.

The figure below depicts the more.

According to the question;

Radius of circular path (r) = 30 m;

Velocity of scooter (v) = 36 km/hr;

= m/s;

= 10 m/s.

Mass of scooter (m) = 150 kg;

For making the turn possible the scooter needs force equal to the centripetal force which is given by;

F =

⇒ F =

⇒ F = 5 x 10 x 10 = 500 N.

The force required is 500 N and it is provided by the normal reaction between the ground and the scooter.

According to the question;

Length of pendulum (l) = 1 m;

Mass of pendulum (m) = 100g =

Velocity at lowest point (v) = 1.4 m/s.

Now;

The Tension in the string is equal to the sum of force in the circular path (centripetal force) and gravitational force..

⇒ Tension T = Centripetal force + gravitational force.

⇒ T = + mg

⇒ T = + 0.1 × 9.8.

(Length of string(l) = radius of circle in which bob of pendulum moves (r))

⇒ T = 0.196 + 0.98

⇒ T = 1.176 N.

The tension in the string is 1.176 N.

The map of India is irregular in shape its center of gravity can be found out in following way

1. Take the map and make three holes along its edge.

2. Suspend the map freely from one hole by using the thread as shown in the figure below.

3. Take a plumb bob and suspend it from the same hole.

4. Draw a line along the thread. It indicates the line of weight at that point. As shown in the figure below.

5. Repeat the same from remaining holes and draw other two lines of weight.

6. The concurrent point of the three lines is the Center of gravity of the map.

Some of the situations where the center of gravity of man lies outside the body are as follows.

i. While doing Yogasanas, if we bend our body in inverted “V” shape. Then the center of gravity may lie out side of the body. The figure below illustrates the point.

ii. The athletes while doing the high jump, bend their body as arc. In that situation, the center of gravity may lie outside of the body. The figure below shows the center of gravity of athlete doing the high jump.

Since the earth is surrounded by the atmosphere with nearly equal thickness and.The center of gravity of Earth lies at the center of Earth because, at that point, the mass of the Earth is equally distributed in all the directions. Therefore center of gravity of the atmosphere of the earth coincides the center of gravity of the earth. As shown in the figure below.

The center of gravity of an object is the point you can suspend the object from without there being any rotation because of the force of gravity .when a boy is doing sit-ups the center of gravity move up and down, thus it is not fixed it keeps changing with the upward and downward movement of the boy. The figure below illustrates it.

The weight vector passes through the base but the move slightly with respect to the previous position.