Beckett Aguirre

2023-03-30

A small rock with mass 0.12 kg is fastened to a massless string with length 0.80 m to form a pendulum. The pendulum is swinging so as to make a maximum angle of 45 with the vertical. Air Resistance is negligible.

a) What is the speed of the rock when the string passesthrough the vertical position?

b) What is the tension in the string when it makes an angle of 45 with the vertical?

c) What is the tension in the string as it passes through the vertical?

Genesis Terrell

Beginner2023-03-31Added 12 answers

To solve this problem, we can use the principles of conservation of energy and the forces acting on the pendulum.

a) To find the speed of the rock when the string passes through the vertical position, we can use the conservation of energy. At the highest point of the swing, the gravitational potential energy is converted entirely into kinetic energy.

The gravitational potential energy at the highest point is given by:

$PE=mgh$,

where $m$ is the mass of the rock, $g$ is the acceleration due to gravity, and $h$ is the height of the rock at the highest point. Since the rock passes through the vertical position, $h$ is equal to the length of the string.

The kinetic energy at the highest point is given by:

$KE=\frac{1}{2}m{v}^{2}$,

where $v$ is the speed of the rock.

According to the conservation of energy, the total mechanical energy remains constant, so we have:

$PE=KE$.

Substituting the expressions for potential and kinetic energy:

$mgh=\frac{1}{2}m{v}^{2}$.

The mass, $m$, cancels out, and we can solve for the speed, $v$:

$v=\sqrt{2gh}$.

Given:

$m=0.12$ kg (mass of the rock)

$g=9.8$ m/s^2 (acceleration due to gravity)

$h=0.80$ m (height of the rock at the highest point)

Substituting these values into the equation:

$v=\sqrt{2\xb79.8\xb70.80}$.

Calculating the value:

$v\approx 2.14$ m/s.

Therefore, the speed of the rock when the string passes through the vertical position is approximately 2.14 m/s.

b) To find the tension in the string when it makes an angle of 45 degrees with the vertical, we need to consider the forces acting on the pendulum at that point.

The tension, $T$, and the weight, $mg$, are the two forces acting on the rock. The vertical component of the tension balances the weight, and the horizontal component provides the centripetal force.

The vertical component of the tension is given by:

$T\mathrm{cos}(\theta )=mg$,

where $\theta $ is the angle with the vertical (45 degrees) and $m$ is the mass of the rock.

The tension can be expressed as:

$T=\frac{mg}{\mathrm{cos}(\theta )}$.

Substituting the values:

$T=\frac{0.12\xb79.8}{\mathrm{cos}({45}^{\circ})}$.

Using the value of $\mathrm{cos}({45}^{\circ})=\frac{\sqrt{2}}{2}$:

$T=\frac{0.12\xb79.8}{\frac{\sqrt{2}}{2}}$.

Calculating the value:

$T\approx 0.83$ N.

Therefore, the tension in the string when it makes an angle of 45 degrees with the vertical is approximately 0.83 N.

c) To find the tension in the string as it passes through the vertical, we need to consider the forces acting on the rock at that point.

At the bottom of the swing, the tension, $T$, provides the centripetal force required to keep the rock moving in a circular path.

The tension can be expressed as:

$T=\frac{m{v}^{2}}{r}$,

where $m$ is the mass of the rock, $v$ is the speed of the rock, and $r$ is the radius of the

a) To find the speed of the rock when the string passes through the vertical position, we can use the conservation of energy. At the highest point of the swing, the gravitational potential energy is converted entirely into kinetic energy.

The gravitational potential energy at the highest point is given by:

$PE=mgh$,

where $m$ is the mass of the rock, $g$ is the acceleration due to gravity, and $h$ is the height of the rock at the highest point. Since the rock passes through the vertical position, $h$ is equal to the length of the string.

The kinetic energy at the highest point is given by:

$KE=\frac{1}{2}m{v}^{2}$,

where $v$ is the speed of the rock.

According to the conservation of energy, the total mechanical energy remains constant, so we have:

$PE=KE$.

Substituting the expressions for potential and kinetic energy:

$mgh=\frac{1}{2}m{v}^{2}$.

The mass, $m$, cancels out, and we can solve for the speed, $v$:

$v=\sqrt{2gh}$.

Given:

$m=0.12$ kg (mass of the rock)

$g=9.8$ m/s^2 (acceleration due to gravity)

$h=0.80$ m (height of the rock at the highest point)

Substituting these values into the equation:

$v=\sqrt{2\xb79.8\xb70.80}$.

Calculating the value:

$v\approx 2.14$ m/s.

Therefore, the speed of the rock when the string passes through the vertical position is approximately 2.14 m/s.

b) To find the tension in the string when it makes an angle of 45 degrees with the vertical, we need to consider the forces acting on the pendulum at that point.

The tension, $T$, and the weight, $mg$, are the two forces acting on the rock. The vertical component of the tension balances the weight, and the horizontal component provides the centripetal force.

The vertical component of the tension is given by:

$T\mathrm{cos}(\theta )=mg$,

where $\theta $ is the angle with the vertical (45 degrees) and $m$ is the mass of the rock.

The tension can be expressed as:

$T=\frac{mg}{\mathrm{cos}(\theta )}$.

Substituting the values:

$T=\frac{0.12\xb79.8}{\mathrm{cos}({45}^{\circ})}$.

Using the value of $\mathrm{cos}({45}^{\circ})=\frac{\sqrt{2}}{2}$:

$T=\frac{0.12\xb79.8}{\frac{\sqrt{2}}{2}}$.

Calculating the value:

$T\approx 0.83$ N.

Therefore, the tension in the string when it makes an angle of 45 degrees with the vertical is approximately 0.83 N.

c) To find the tension in the string as it passes through the vertical, we need to consider the forces acting on the rock at that point.

At the bottom of the swing, the tension, $T$, provides the centripetal force required to keep the rock moving in a circular path.

The tension can be expressed as:

$T=\frac{m{v}^{2}}{r}$,

where $m$ is the mass of the rock, $v$ is the speed of the rock, and $r$ is the radius of the

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