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4 years ago

14

A fidget spinner experiences a constant torque of 1.4 N.m. If the spinner is initially at rest, what is its angular momentum 2.0

s later?

1 answer:

motikmotik4 years ago

4 0

Answer:

$2.8N-msec$

Explanation:

We have given torque $\tau =1.4N-m$

Initial time $t_1=0\ sec$

Final time $t_2=2\ sec$

There is relation between angular momentum and torque

that is $\frac{dL}{dt}=\tau$

$dL=\tau dt$

$\int dL=\tau \oint_{t_1}^{t_2}dt$

$L=\tau (t_1-t_2)=1.4\times (2-0)=2.8N-msec$

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AysviL [449]

Answer:

The impulse transferred to the nail is 0.01 kg*m/s.

Explanation:

The impulse (J) transferred to the nail can be found using the following equation:

$J = \Delta p = p_{f} - p_{i}$

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$p_{f}$ : is the final momentum

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$p_{i} = m_{1}v_{1_{i}} + m_{2}v_{2_{i}}$

Where 1 is for the hammer and 2 is for the nail.

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$p_{i} = m_{1}v_{1_{i}} = 0.25 kg*(-10 m/s) = -2.5 kg*m/s$

Now, the final momentum can be found taking into account that the hammer remains in contact with the nail during and after the blow:

$p_{f} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}}$

Since the hammer and the nail are moving in the negative direction:

$v_{1_{f}}$ = $v_{2_{f}}$ = -9.7 m/s

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$J = p_{f} - p_{i} = - 2.49 kg*m/s + 2.50 kg*m/s = 0.01 kg*m/s$

Therefore, the impulse transferred to the nail is 0.01 kg*m/s.

I hope it helps you!

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