Question

A girl on a merry-go-round platform holds a pendulum in her hand. The pendulum is \(6.00 \mathrm{~m}\) from the rotation axis of the platform. The rotational speed of the platform is 0.0200 rev/s. It is found that the pendulum hangs at an angle \(\theta\) to the vertical. Find $\theta

Short answer

Answer: The angle between the pendulum and the vertical direction is approximately 10.17°.

Step-by-step solution

Identify the forces acting on the pendulum bob

The forces acting on the pendulum bob are the gravitational force (mg) acting vertically downward and the centripetal force (mv^2/r) acting horizontally along the radial direction.

Resolve the gravitational force into radial and tangential components

In order to analyze the forces acting on the bob, we need to break down the gravitational force into two components: one is radial (along the radial direction) and the other is tangential (perpendicular to the radial direction). This can be done using the angle θ, where the radial component (F_r) is given by mgcosθ, and the tangential component (F_t) is given by mgsinθ.

Formulate the equation for the centripetal force

The centripetal force acting on the pendulum bob due to the rotation of the platform is mv^2/r, where m is the mass of the bob, v is the linear velocity due to the rotation, and r is the distance from the rotation axis (given as 6.00 m). We can obtain v by multiplying the rotational speed (ω) and r. So, v = ωr = (0.0200 rev/s * 6.00 m) * (2π rad/rev) = 0.754 rad*m/s.

Equate the radial component of gravitational force to centripetal force

Since the pendulum is in equilibrium, the radial component of gravitational force should be equal to the centripetal force acting on the pendulum bob. Hence, we have mgcosθ = mv^2/r. Now we can plug in the given values and solve for θ: cosθ = (mv^2/r) / mg. We can cancel the mass, so cosθ = (0.754^2 rad^2*m^2/s^2) / (6.00m * 9.81m/s^2).

Calculate the angle θ between the pendulum and the vertical direction

Now, solving for θ, we get θ = arccos(cosθ) = arccos((0.754^2 rad^2*m^2/s^2) / (6.00m * 9.81m/s^2)) = 10.17°. Hence, the angle between the pendulum and the vertical direction is approximately 10.17°.

Key concepts

Centripetal Force

One of the fundamental forces in merry-go-round physics is the centripetal force, which keeps objects moving in a circle. This isn't a type of force generated by a unique source, rather it's the net result of forces acting on the object to keep it in circular motion. For instance, the tension in a string or the friction between car tires and the road can provide the necessary centripetal force to maintain circular movement.

In our merry-go-round example, a girl is holding a pendulum that experiences centripetal force due to the platform’s rotation. The equation for centripetal force is derived from Newton's second law, and is given by \( F_c = \frac{mv^2}{r} \), where \( m \) is the object's mass, \( v \) is its velocity, and \( r \) is the radius of the circle.

As the merry-go-round spins, centripetal force acts horizontally, pulling the pendulum towards the center. This is why the pendulum bob hangs at an angle from the vertical; it's the balance point where the centripetal force is equivalent to the radial component of the pendulum's weight.

Rotational Motion

In merry-go-rounds and other systems involving rotational motion, it's important to understand how objects move along a curved path. Unlike linear motion, rotational motion deals with angular velocity and acceleration. The speed at which the object moves around the circle is represented by its angular velocity \( \omega \). For our example, the platform has an angular velocity of 0.0200 revolutions per second (rev/s).

To find the linear velocity \( v \) from angular velocity, we use the relationship \( v = \omega r \), where \( r \) is the radius from the center of rotation to the object. This formula allows us to convert the merry-go-round’s rotational speed into linear velocity to further analyze the pendulum's motion.

Understanding rotational motion is crucial for calculating the forces at play because the motion dictates the magnitude of the centripetal force required to keep objects moving in their circular path.

Pendulum Motion

A pendulum in motion, such as in our example of a girl on a merry-go-round, combines simple harmonic motion with circular motion principles. Even though the platform is spinning, the pendulum bob wants to hang vertically due to gravity. However, because of the circular rotation, it’s forced to hang at an angle. This setup creates pendulum motion on a merry-go-round which is complex as compared to the regular vertical motion of a typical pendulum.

The force acting on the pendulum bob has two components: a tangential component along the path of the pendulum's swing, and a radial component that points towards the center of the circular path. The pendulum motion becomes a blend of swinging due to gravity and the forced circular path due to the rotation of the merry-go-round.

When dissecting the forces acting on the pendulum, the angle \( \theta \) that it forms with the vertical is determined by the equilibrium between gravitational force pulling it down and the centripetal force pulling it towards the center of the merry-go-round’s rotation. Therefore, understanding both pendulum and circular motion is essential to solving physics problems involving pendulums in non-traditional settings.