Shortly after a vinyl record (radius 0.152 m) starts rotating on a turntable, its angular velocity is 1.60 rad/s and increasing at a rate of 8.00 rad/s2. At this instant, for a point at the rim of the record, what is the tangential component of acceleration

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Answer:

The tangential component of acceleration is 1.216 [tex]\frac{m}{s^{2} }[/tex]

Explanation:

Given:

Radius of turntable [tex]r = 0.152[/tex] m

Angular velocity [tex]\omega = 1.60[/tex] [tex]\frac{rad}{s}[/tex]

Angular acceleration [tex]\alpha = 8[/tex] [tex]\frac{rad}{s^{2} }[/tex]

For finding tangential component of acceleration,

We use rotational mechanics,

Tangential acceleration is given by,

  [tex]a_{tan} = \alpha r[/tex]

Where [tex]\alpha =[/tex] angular acceleration, [tex]r =[/tex] radius of turntable

  [tex]a_{tan} = 8 \times 0.152[/tex]

  [tex]a_{tan} = 1.216[/tex] [tex]\frac{m}{s^{2} }[/tex]

Therefore, the tangential component of acceleration is 1.216 [tex]\frac{m}{s^{2} }[/tex]

The tangential component of acceleration will be 1.216 m/s².The acceleration has two components tangential as well as radial acceleration.

What is acceleration?

Acceleration is defined as the rate of change of the velocity of the body. Its unit is m/sec².It is a vector quantity. It requires both magnitudes as well as direction to define.

The given data in the problem is

r is the radius=  0.152 m

[tex]\rm \omega[/tex] is the angular velocity = 1.60 rad/s

[tex]\rm \alpha[/tex] is the angular acceleration = 8.00 rad/s².

The tangential component of acceleration = ?

The tangential acceleration is found by the formula s;

[tex]\rm a_{tan} = r \alpha \\\\ \rm a_{tan} = 8 \times 0.152 \\\\ \rm a_{tan} =1.216\ m/s^2[/tex]

Hence the tangential component of acceleration will be 1.216 m/s².

To learn more about acceleration refer to the link;

https://brainly.com/question/969842