The bigclaw snapping shrimp shown in (Figure 1) is aptly named–it has one big claw that snaps shut with remarkable speed. The part of the c

Question

The bigclaw snapping shrimp shown in (Figure 1) is aptly named–it has one big claw that snaps shut with remarkable speed. The part of the c

Question

The bigclaw snapping shrimp shown in (Figure 1) is aptly named–it has one big claw that snaps shut with remarkable speed. The part of the claw that moves rotates through a 90∘ angle in 1.3 ms. We assume that the claw is 1.5 cm long and that it undergoes a constant angular acceleration. (you don’t need figure to solve problem)

1. What is the angular acceleration? Assume the claw starts from rest.
Express your answer in radians per second squared.

2. What is the final angular speed of the claw?
Express your answer in radians per second.

3. What is the tangential acceleration of the tip of the claw?
Express your answer with the appropriate units.

4. How fast is the tip of the claw moving at the end of its motion?
Express your answer with the appropriate units.

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Physics Gerda 4 years 2021-07-23T10:09:41+00:00 2021-07-23T10:09:41+00:00 1 Answers 608 views 0

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Dulcie · Answer 1 · 2021-07-23T10:10:59+00:00

1) $1.86\cdot 10^6 rad/s^2$

2) 2418 rad/s

3) $27000 m/s^2$

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:

$\theta=\omega_i t +\frac{1}{2}\alpha t^2$

where:

$\theta$ is the angular displacement

$\omega_i$ is the initial angular velocity

t is the time

$\alpha$ is the angular acceleration

In this problem we have:

$\theta=90^{\circ} = \frac{\pi}{2}rad$ is the angular displacement

t = 1.3 ms = 0.0013 s is the time elapsed

$\omega_i = 0$ is the initial angular velocity

Solving for $\alpha$ , we find:

$\alpha = \frac{2(\theta-\omega_i t)}{t^2}=\frac{2(\pi/2)-0}{0.0013}=1.86\cdot 10^6 rad/s^2$

2)

For an object in accelerated rotational motion, the final angular speed can be found by using another suvat equation:

$\omega_f = \omega_i + \alpha t$

where

$\omega_i$ is the initial angular velocity

t is the time

$\alpha$ is the angular acceleration

In this problem we have:

t = 1.3 ms = 0.0013 s is the time elapsed

$\omega_i = 0$ is the initial angular velocity

$\alpha = 1.86\cdot 10^6 rad/s$ is the angular acceleration

Therefore, the final angular speed is:

$\omega_f = 0 + (1.86\cdot 10^6)(0.0013)=2418 rad/s$

3)

The tangential acceleration is related to the angular acceleration by the following formula:

$a_t = \alpha r$

where

$a_t$ is the tangential acceleration

$\alpha$ 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:

$\alpha = 1.86\cdot 10^6 rad/s$ 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,

$a_t=(1.86\cdot 10^6)(0.015)=27900 m/s^2$

4)

Since the tip of the claw is moving by uniformly accelerated motion, we can find its final speed using the suvat equation:

$v=u+at$

where

u is the initial linear speed

a is the tangential acceleration

t is the time elapsed

Here we have:

$a=27900 m/s^2$ (tangential acceleration)

u = 0 m/s (it starts from rest)

t = 1.3 ms = 0.0013 s is the time elapsed

Substituting,

$v=0+(27900)(0.0013)=36.3 m/s$

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The bigclaw snapping shrimp shown in (Figure 1) is aptly named–it has one big claw that snaps shut with remarkable speed. The part of the c