In 1980, over San Francisco Bay, a large yo-yo was released from a crane. Suppose the yo-yo was 153 kg, and it consisted of two uniform disks of radius 35.4 cm connected by an axle of radius 3.19 cm. What was the magnitude of the acceleration of the yo-yo during (a) its fall and (b) its rise? (c)What was the tension in the cord on which it rolled?
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First two parts, the answer is in m/s^2
Third is in N
thanks
find the moment of inertia for the two disks (each has half the mass).
tension*radius= momentinertia*angular acceleartion
and tension= mg-ma=mg-m*angacceleration*r
so you have two equations, two unknowns, solve for tension, and angular acceleartion.
on the way up,
tension*radius=momentinertia*angularacceleration
tension= mg+ma=mg+m(angaccleration*r
The mass or moment of the axle must be specified because its moment of inertia affects the answer.
To find the magnitude of the acceleration of the yo-yo during its fall and rise, we can use the principle of conservation of energy.
(a) During the fall of the yo-yo, the total mechanical energy is transformed from gravitational potential energy to rotational kinetic energy as it descends. The formula to calculate the acceleration during this phase is:
Acceleration = (2 * g * h) / (r₁² + r₂²),
where g is the acceleration due to gravity (approximately 9.8 m/s²), h is the height from which the yo-yo is released, r₁ is the radius of the larger disk, and r₂ is the radius of the axle.
Given that the radius of the larger disk is 35.4 cm (or 0.354 m), and the radius of the axle is 3.19 cm (or 0.0319 m), we can substitute these values into the acceleration formula:
Acceleration = (2 * 9.8 * h) / (0.354² + 0.0319²)
(b) During the rise of the yo-yo, the total mechanical energy is transformed from rotational kinetic energy back to gravitational potential energy. The formula to calculate the acceleration during this phase is the same as in part (a).
Acceleration = (2 * g * h) / (r₁² + r₂²)
(c) To determine the tension in the cord on which the yo-yo rolled, we can use the centripetal force formula:
Tension = m * a,
where m is the mass of the yo-yo (153 kg) and a is the acceleration calculated in part (a) or (b).
After calculating the acceleration in parts (a) and (b), substitute the values in the tension formula to find the tension in the cord.
To find the magnitude of acceleration during the fall and rise of the yo-yo as well as the tension in the cord, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration (F = m × a).
(a) Acceleration during the fall:
To calculate the acceleration of the yo-yo during its fall, we need to consider the forces acting on it. The main forces are the force of gravity pulling it down and the tension in the cord pulling it up. Since the yo-yo is in free fall, the tension in the cord can be neglected. Therefore, the net force is simply the force of gravity.
The force of gravity (Fg) can be calculated using the formula Fg = m × g, where m is the mass of the yo-yo and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Fg = 153 kg × 9.8 m/s^2
Fg = 1499.4 N
Since the yo-yo consists of two disks connected by an axle, the net force acting on it is equal to the product of its total mass and acceleration:
Fnet = m × a
1499.4 N = 153 kg × a
To find the magnitude of acceleration (a), divide both sides of the equation by the mass:
a = 1499.4 N / 153 kg
a ≈ 9.8 m/s^2
Therefore, the magnitude of the acceleration of the yo-yo during its fall is approximately 9.8 m/s^2.
(b) Acceleration during the rise:
During the rise of the yo-yo, the forces acting on it are similar to the fall but in the opposite direction. In this case, the tension in the cord will play a significant role.
The net force acting on the yo-yo during its rise is given by the difference between the force of gravity and the tension in the cord:
Fnet = Fg - T
Using the same values as before:
Fnet = 1499.4 N - T
Since the yo-yo is moving upward, the net force will be in the opposite direction of the acceleration during the fall. Therefore, the net force is negative:
Fnet = -153 kg × a
Setting the two equations equal to each other:
1499.4 N - T = -153 kg × a
Now, we can substitute the value of acceleration (a) as -9.8 m/s^2 (opposite in direction to the acceleration during the fall):
1499.4 N - T = -153 kg × (-9.8 m/s^2)
Simplifying the equation:
1499.4 N - T = 1500.6 N
To find the tension in the cord (T), subtract 1499.4 N from both sides:
T = 1500.6 N - 1499.4 N
T ≈ 1.2 N
Therefore, the tension in the cord during the rise of the yo-yo is approximately 1.2 N.
(c) Tension in the cord:
As calculated in part (b), the tension in the cord during the rise of the yo-yo is approximately 1.2 N.