A 67.5 kg hoop rolls along a horizontal floor so that its center of mass has a speed of 0.090 m/s. How much work must be done to stop it?
To calculate the work required to stop the hoop, we first need to understand the concept of work and its formula.
Work, denoted by the symbol "W," is the energy transfer that occurs when a force is applied to an object, causing it to move. It is calculated using the formula:
W = F * d * cos(theta)
Where:
- W is the work done (in joules, J)
- F is the force applied (in newtons, N)
- d is the displacement of the object (in meters, m)
- theta is the angle between the force and the displacement vectors (in degrees)
In this case, we need to stop the hoop, which means we need to bring it to rest. Therefore, we want the final velocity of the hoop to be zero.
When the hoop is rolling, it possesses both linear and rotational kinetic energy. To bring the hoop to rest, the work must be done both against its linear kinetic energy and its rotational kinetic energy.
The linear kinetic energy (K_linear) of an object can be calculated using the formula:
K_linear = 0.5 * m * v^2
Where:
- K_linear is the linear kinetic energy (in joules, J)
- m is the mass of the object (in kilograms, kg)
- v is the velocity of the object (in meters per second, m/s)
Given that the hoop has a mass of 67.5 kg and a velocity of 0.090 m/s, we can calculate its linear kinetic energy.
K_linear = 0.5 * 67.5 kg * (0.090 m/s)^2
K_linear = 0.273375 J
Next, we need to calculate the rotational kinetic energy (K_rotational) of the hoop. The rotational kinetic energy depends on the moment of inertia (I) and the angular velocity (ω) of the object:
K_rotational = 0.5 * I * ω^2
Since this problem provides no information about the hoop's dimensions, we can assume it is a solid hoop. The moment of inertia of a solid hoop about its center is given by the formula:
I = m * r^2
Where:
- I is the moment of inertia (in kilogram-meter squared, kg·m^2)
- m is the mass of the hoop (in kilograms, kg)
- r is the radius of the hoop (in meters, m)
A solid hoop's moment of inertia can be calculated by substituting the values of mass and radius into the equation. Assuming a standard hoop, the moment of inertia can be calculated as:
I = m * r^2
I = 67.5 kg * (r^2)
The radius of a hoop is the distance from its center of mass to any point on the edge, which is given by the formula:
r = (1/2) * d
Where:
- r is the radius of the hoop (in meters, m)
- d is the diameter of the hoop (in meters, m)
The diameter of a hoop is twice its radius:
d = 2 * r
Substituting this equation into the previous one, we obtain:
r = (1/2) * (2 * r)
r = r
Hence, the radius (r) of the hoop is equal to its radius. Therefore, we can replace "r" in the equation for the moment of inertia (I) with "r" in this context.
I = m * r^2
I = 67.5 kg * (r^2)
Finally, we can calculate the rotational kinetic energy.
K_rotational = 0.5 * I * ω^2
K_rotational = 0.5 * (67.5 kg * (r^2)) * (v/r)^2
K_rotational = 0.5 * 67.5 kg * v^2
Now, we have both the linear kinetic energy and the rotational kinetic energy values. To find the total work done to stop the hoop, we sum these two energies:
Total work = K_linear + K_rotational
Total work = 0.273375 J + (0.5 * 67.5 kg * (0.090 m/s)^2)
Calculating this expression will give us the answer, which is the total work required to stop the hoop.