A 60.0-kg man jumps 1.70 m down onto a concrete walkway. His downward motion stops in 0.025 seconds. If he forgets to bend his k
1 answer:
Answer:
The magnitude of the force transmitted to his leg bones is 1413.6 N
Explanation:
Recall that force is defined as the change in linear momentum per unit time:
We can use this formula to find the force transmitted to his legs. We know that the final momentum () is 0 since the person is not moving on the floor, but we need to find what the person's momentum was an instant before he touches the ground. Since we know he person's mass, all we need for the initial momentum is his velocity.
For such we use conservation of energy in free fall, knowing that he jumped from 1.7 meters:
Now with this velocity, we know the (initial momentum) just before impact.
And since the impact lasted 0.025 seconds, we can find the force using the first formula we recalled:
so the magnitude of the force is 1413.6 N
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Answer:
a) v = 1.01 m/s
b) a = 5.6 m/s²
Explanation:
a)
- If the disk is initially at rest, and it is applied a constant force tangential to the rim, we can apply the following expression (that resembles Newton's 2nd law, applying to rigid bodies instead of point masses) as follows:
- Where τ is the external torque applied to the body, I is the rotational inertia of the body regarding the axis of rotation, and α is the angular acceleration as a consequence of the torque.
- Since the force is applied tangentially to the rim of the disk, it's perpendicular to the radius, so the torque can be calculated simply as follows:
- τ = F*r (2)
- For a solid uniform disk, the rotational inertia regarding an axle passing through its center is just I = m*r²/2 (3).
- Replacing (2) and (3) in (1), we can solve for α, as follows:
- Since the angular acceleration is constant, we can use the following kinematic equation:
- Prior to solve it, we need to convert the angle rotated from revs to radians, as follows:
- Replacing (6) in (5), taking into account that ω₀ = 0 (due to the disk starts from rest), we can solve for ωf, as follows:
- Now, we know that there exists a fixed relationship the tangential speed and the angular speed, as follows:
- where r is the radius of the circular movement. If we want to know the tangential speed of a point located on the rim of the disk, r becomes the radius of the disk, 0.200 m.
- Replacing this value and (7) in (8), we get:
b)
- There exists a fixed relationship between the tangential and the angular acceleration in a circular movement, as follows:
- where r is the radius of the circular movement. In this case the point is located on the rim of the disk, so r becomes the radius of the disk.
- Replacing this value and (4), in (9), we get:
- Now, the resultant acceleration of a point of the rim, in magnitude, is the vector sum of the tangential acceleration and the radial acceleration.
- The radial acceleration is just the centripetal acceleration, that can be expressed as follows:
- Since we are asked to get the acceleration after the disk has rotated 0.2 rev, and we have just got the value of the angular speed after rotating this same angle, we can replace (7) in (11).
- Since the point is located on the rim of the disk, r becomes simply the radius of the disk,, 0.200 m.
- Replacing this value and (7) in (11) we get:
- The magnitude of the resultant acceleration will be simply the vector sum of the tangential and the radial acceleration.
- Since both are perpendicular each other, we can find the resultant acceleration applying the Pythagorean Theorem to both perpendicular components, as follows: