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hammer [34]
2 years ago
5

A wheel of radius 0.5 m rotates with a constant angular speed about an axis perpendicular to its center. A point on the wheel th

at is 0.2 m from the center has a tangential speed of 2 m/s. Determine the tangential acceleration of the point that is 0.2 m from the center.
Physics
1 answer:
Masja [62]2 years ago
8 0

Answer:

The tangential acceleration is 0 m/s².

Explanation:

Given:

Radius of the wheel = 0.5 m

The point of observation for calculating tangential acceleration = 0.2 m from center.

Tangential speed at the point of observation = 2 m/s

The angular speed of the wheel is a constant.

In order to determine the tangential acceleration, we make use of the following formula:

Tangential acceleration at a point = Angular acceleration × Distance of the point from center

Or, a_t=\alpha \times r

Now, angular acceleration is defined as the rate of change of angular speed.

Here, the angular speed of the wheel is a constant. So, the change of angular speed is 0. Therefore, the angular acceleration is also 0 rad/s².

Now, from the above formula, as angular acceleration is 0, the magnitude of tangential acceleration at a point that is 0.2 m from the center of the wheel is also 0 m/s².

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Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m
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Explanation:

In the image  attached with this answer are shown the given options from which only one is correct.

The correct expression is:

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Because, if we derive velocity v_{t} with respect to time t we will have acceleration a, hence:

m \frac{d}{dt}v_{(t)}=m.a

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The other expressions are incorrect, let’s prove it:

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m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m This result has units of kg

m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C This result has units of kgm^{2} and C is a constant

m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m This result has units of kg

m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m This result has units of kg

\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C This result has units of kg \frac{m^{3}}{s^{3}} and C is a constant

m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C This result has units of kg \frac{m^{2}}{s^{4}} and C is a constant

\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0 because v_{(x)} is a constant in this derivation respect to t

m\int v_{(t)} dt= \frac{m {v_{(t)}}^{2}}{2}+C This result has units of kg \frac{m^{2}}{s^{2}} and C is a constant

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Answer:

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1200.7 + M_1 = 1518.4

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