# A small 0.18 kg metal ball is tied to a very light (essentially massless) string that is 0.8 m long. The string is attached to the ceiling so as to form a pendulum. The pendulum is set into motion by releasing it from rest at an angle of 40 ∘ with the vertical.

What is the speed of the ball when it reaches the bottom of the arc? What is the centripetal acceleration of the ball at this point?What is the tension in the string at this point?

## height above bottom at 40deg

h = .8(1-cos 40)

compute h

potential energy at that height

U = m g h = .18*9.81*h

compute U

so

kinetic energy at bottom Ke = (1/2)m v^2 = U

compute v

Ac = v^2/R = v^2/.8

compute Ac

T = m g + m Ac

## Why did the metal ball go to the party? Because it wanted to swing into action!

To find the speed of the ball at the bottom of the arc, we can use the principle of conservation of energy. At the topmost point of the arc, all the potential energy is converted to kinetic energy. So, the potential energy at that point is given by the formula: PE = mgh, where m is the mass of the ball, g is the acceleration due to gravity, and h is the vertical height at the topmost point of the arc.

Using the given values, the potential energy at the top is: PE = (0.18 kg)(9.8 m/s^2)(0.8 m)(cos 40°). But since the kinetic energy is maximum when the ball is at the bottom of the arc, the potential energy at the top is equal to the kinetic energy at the bottom: KE = (0.18 kg)(9.8 m/s^2)(0.8 m)(cos 40°).

To calculate the speed, we can use the formula: KE = 1/2 mv^2, where m is the mass of the ball and v is the speed at the bottom. Rearranging the equation to solve for v, we get: v = sqrt((2 KE) / m).

By substituting the values, we find: v = sqrt((2 (0.18 kg)(9.8 m/s^2)(0.8 m)(cos 40°)) / 0.18 kg).

As for the centripetal acceleration at the bottom of the arc, we can use the formula: ac = v^2 / r, where v is the speed at the bottom and r is the length of the string.

By substituting the values, we find: ac = (v^2) / 0.8 m.

The tension in the string at the bottom of the arc can be determined using the formula: T = mg + ma, where m is the mass of the ball, g is the acceleration due to gravity, and a is the centripetal acceleration at the bottom.

By substituting the values, we find: T = (0.18 kg)(9.8 m/s^2) + (0.18 kg)(v^2) / 0.8 m.

Just remember, these calculations assume that there's no friction or other external forces acting on the ball during its motion. Now, if only we could calculate the tension in social situations as easily as we calculate the tension in strings...

## To find the speed of the ball when it reaches the bottom of the arc, we can use the principle of conservation of mechanical energy.

1. Potential Energy at the top: The potential energy at the top of the arc is equal to the potential energy at the bottom (when the ball reaches its maximum speed). Using the formula for potential energy, we have:

mgh = (1/2)mv^2

2. Solving for v, we can eliminate m:

gh = (1/2)v^2

3. Now, we can substitute the given values: g = 9.8 m/s^2, and h = (0.8 m) * (1 - cos(40°)).

We can use the identity cos(2θ) = 1 - 2sin^2(θ), and in our case, θ = 40°.

cos(80°) = 1 - 2sin^2(40°)

cos(80°) = 1 - 2(1 - cos^2(40°))

cos(80°) = 3cos^2(40°) - 2

4. Rearranging the equation, we have:

cos^2(40°) = (1 + cos(80°)) / 3

cos(40°) ≈ 0.766

5. Now, we can calculate the height:

h = (0.8 m) * (1 - 0.766) ≈ 0.188 m

6. Substituting the values into the equation from step 2, we have:

(9.8 m/s^2)(0.188 m) = (1/2)v^2

7. Solving for v:

v^2 ≈ 1.8444

v ≈ √1.8444

v ≈ 1.357 m/s

Therefore, the speed of the ball when it reaches the bottom of the arc is approximately 1.357 m/s.

Now, let's find the centripetal acceleration of the ball at this point:

1. The centripetal acceleration can be calculated using the equation:

a = v^2 / r

2. Substituting the values, we have:

a = (1.357 m/s)^2 / 0.8 m

3. Solving for a:

a ≈ 2.301 m/s^2

Therefore, the centripetal acceleration of the ball at the bottom of the arc is approximately 2.301 m/s^2.

Finally, let's calculate the tension in the string at this point:

1. The tension in the string can be found by considering the forces acting on the ball at the bottom of the arc. The tension in the string provides the centripetal force necessary to keep the ball moving in a circle.

2. Using Newton's second law, we can write:

T - mg = ma

3. Substituting the values, we have:

T - (0.18 kg)(9.8 m/s^2) = (0.18 kg)(2.301 m/s^2)

4. Solving for T:

T ≈ 2.63 N

Therefore, the tension in the string at the bottom of the arc is approximately 2.63 N.

## To find the speed of the ball when it reaches the bottom of the arc, we can use the conservation of mechanical energy. At the highest point of the swing, all of the potential energy is converted into kinetic energy at the bottom. Therefore, the initial potential energy at the highest point is equal to the kinetic energy at the lowest point.

First, let's calculate the height of the ball from the lowest point:

h = L(1 - cosθ)

where h is the height, L is the length of the string, and θ is the angle with the vertical.

Plugging in the values:

h = 0.8 m (1 - cos(40°))

h = 0.8 m (1 - 0.766)

h = 0.8 m (0.234)

h ≈ 0.187 m

Now we can calculate the potential energy at the highest point:

PE = mgh

where PE is the potential energy, m is the mass, g is the acceleration due to gravity, and h is the height.

Plugging in the values:

PE = (0.18 kg)(9.8 m/s^2)(0.187 m)

PE ≈ 0.328 J

As the ball reaches the bottom of the arc, all of the potential energy is converted into kinetic energy. Therefore, the kinetic energy at the lowest point is equal to the potential energy at the highest point:

KE = PE

Using the formula for kinetic energy:

KE = 0.5mv^2

where KE is the kinetic energy and v is the velocity.

Rearranging the equation:

v = sqrt((2KE) / m)

Plugging in the values:

v = sqrt((2 * 0.328 J) / 0.18 kg)

v ≈ sqrt(3.64 m^2/s^2)

v ≈ 1.91 m/s

So, the speed of the ball when it reaches the bottom of the arc is approximately 1.91 m/s.

To find the centripetal acceleration of the ball at the bottom of the arc, we can use the equation:

a = v^2 / r

where a is the acceleration, v is the velocity, and r is the radius (length of the string).

Plugging in the values:

a = (1.91 m/s)^2 / 0.8 m

a ≈ 4.58 m/s^2

So, the centripetal acceleration of the ball at the bottom of the arc is approximately 4.58 m/s^2.

To find the tension in the string at the bottom of the arc, we can use the equation for centripetal force:

F = m * a

where F is the force, m is the mass, and a is the centripetal acceleration.

Plugging in the values:

F = (0.18 kg) * (4.58 m/s^2)

F ≈ 0.824 N

Therefore, the tension in the string at the bottom of the arc is approximately 0.824 N.