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2 years ago

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A wheel rotating about a fixed axis has a constant angular acceleration of 4.0 rad/s2. In a 4.0-s interval the wheel turns throu

gh an angle of 80 radians. Assuming the wheel started from rest, how long had it been in motion at the start of the 4.0-s interval?
a. 3 s.
b. 6 s.
c. 9 s.
d. 12 s.

1 answer:

anastassius [24]2 years ago

8 0

Answer:

a. 3 s.

Explanation:

Given;

angular acceleration of the wheel, α = 4 rad/s²

time of wheel rotation, t = 4 s

angle of rotation, θ = 80 radians

Apply the kinematic equation below,

$\theta = \omega_1 t \ + \ \frac{1}{2} \alpha t^2\\\\80 = 4\omega_1 + \frac{1}{2}*4*4^2\\\\80 = 4\omega_1 + 32\\\\ 4\omega_1 = 48\\\\ \omega_1 = \frac{48}{4}\\\\ \omega_1 = 12 \ rad/s$

Given initial angular velocity, ω₀ = 0

Apply the kinematic equation below;

$\omega_1 = \omega_o + \alpha t_1\\\\12 = 0 + 4t\\\\4t = 12\\\\t = \frac{12}{4}\\\\t = 3 \ s$

Therefore, the wheel had been in motion for 3 seconds.

a. 3 s.

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Part(A): The magnitude of Jill's final velocity is $\bf{6.59~m/s}$ .

Part(B): The direction is $\bf{42.7^{0}}$ south to east.

Explanation:

Given:

Mass of Jack, $m_{1} = 59.0~Kg$

Mass of Jill, $m_{2} = 47..0~Kg$

Initial velocity of Jack, $v_{1i} = 8.00~m/s$

Initial velocity of Jill, $v_{2i} = 0$

Final velocity of Jack, $v_{1f} 5.00~m/s$

The final angle made by Jack after collision, $\alpha = 34.0^{0}$

Consider that the final velocity of Jill be $v_{2f}$ and it makes an angle of $\beta$ with respect to east, as shown in the figure.

Conservation of momentum of the system along east direction is given by

$~~~~&& m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f} \cos \alpha + m_{2}v_{2f}^{x}\\&or,& v_{2f}^{x} = \dfrac{m_{1}(v_{1i} - v_{1f} \cos \alpha)}{m_{2}}$

where, $v_{2f}^{x}$ is the component of Jill's final velocity along east. The direction of this component will be along east.

Substituting the value, we have

$v_{2f}^{x} &=& \dfrac{(59.0~Kg)(8.00~m/s - 5.00 \cos 34.0^{0}~m/s)}{47.0~Kg}\\~~~~~&=& 4.84~m/s$

Conservation of momentum of the system along north direction is given by

$~~~~&& v_{2f}^{y} + v_{1f} \sin \alpha = 0\\&or,& v_{2f}^{y} = - v_{1f} \sin \alpha = (8.00~m/s) \sin 34^{0} = 4.47~m/s$

where, $v_{2f}^{y}$ is the component of Jill's final velocity along north. The direction of this component will be along the opposite to north.

Part(A):

The magnitude of the final velocity of Jill is given by

$v_{2f} &=& \sqrt{(v_{2f}^{x})^{2} + (v_{2f}^{y})^{2}}\\~~~~~&=& 6.59~m/s$

Part(B):

The direction is given by

$\beta &=& \tan^{-1}(\dfrac{4.47~m/s}{4.84~m/s})\\~~~~&=& 42.7^{0}$

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