Certain insects can achieve seemingly impossible accelerations while jumping. the click beetle accelerates at an astonishing 400
1 answer:
(a) The launching velocity of the beetle is 6.4 m/s
(b) The time taken to achieve the speed for launch is 1.63 ms
(c) The beetle reaches a height of 2.1 m.
(a) The beetle starts from rest and accelerates with an upward acceleration of 400 g and reaches its launching speed in a distance 0.53 cm. Here g is the acceleration due to gravity.
Use the equation of motion,
Here, the initial velocity of the beetle is u, its final velocity is v, the acceleration of the beetle is a, and the beetle accelerates over a distance s.
Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 0.52×10⁻²m for s.
The launching speed of the beetle is <u>6.4 m/s</u>.
(b) To determine the time t taken by the beetle for launching itself upwards is determined by using the equation of motion,
Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 6.385 m/s for v.
The time taken by the beetle to launch itself upwards is <u>1.62 ms</u>.
(c) After the beetle launches itself upwards, it is acted upon by the earth's gravitational force, which pulls it downwards towards the earth with an acceleration equal to the acceleration due to gravity g. Its velocity reduces and when it reaches the maximum height in its path upwards, its final velocity becomes equal to zero.
Use the equation of motion,
Substitute 6.385 m/s for u, -9.8 m/s² for g and 0 m/s for v.
The beetle can jump to a height of <u>2.1 m</u>
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=
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1)
2) 2418 rad/s
3)
4) 36.3 m/s
Explanation:
1)
The angular acceleration of an object in rotation is the rate of change of angular velocity.
It can be calculated using the following suvat equation for angular motion:
where:
is the angular displacement
is the initial angular velocity
t is the time
is the angular acceleration
In this problem we have:
is the angular displacement
t = 1.3 ms = 0.0013 s is the time elapsed
is the initial angular velocity
Solving for , we find:
2)
For an object in accelerated rotational motion, the final angular speed can be found by using another suvat equation:
where
is the initial angular velocity
t is the time
is the angular acceleration
In this problem we have:
t = 1.3 ms = 0.0013 s is the time elapsed
is the initial angular velocity
is the angular acceleration
Therefore, the final angular speed is:
3)
The tangential acceleration is related to the angular acceleration by the following formula:
where
is the tangential acceleration
is the angular acceleration
r is the distance of the point from the centre of rotation
Here we want to find the tangential acceleration of the tip of the claw, so:
is the angular acceleration
r = 1.5 cm = 0.015 m is the distance of the tip of the claw from the axis of rotation
Substituting,
4)
Since the tip of the claw is moving by uniformly accelerated motion, we can find its final speed using the suvat equation:
where
u is the initial linear speed
a is the tangential acceleration
t is the time elapsed
Here we have:
(tangential acceleration)
u = 0 m/s (it starts from rest)
t = 1.3 ms = 0.0013 s is the time elapsed
Substituting,