The outside edge of a spinning compact disc is moving at a higher velocity than an inside track on the disc.
question 3 options:
true
false

The outside edge of a spinning compact disc moves with a higher velocity than the inner track of the disc.

Explanation:

Here the compact disc undergoes rotational motion about a fixed axis which is its centre in this case. The particles in rotational motion have angular velocity which is given by the equation

ω = ∅/t

Where  θ is the angular displacement and t is the time.

The transnational speed of a particle which is in circular motion is given by the equation

v = rω

r is the distance of the point from the rotation centre

The transnational speed of the particles is merely determined by their distance from the centre in this case. It is due to the equality of angular velocity of all the points.

The distance of the outer edge of the compact disc from its rotational centre  is larger than  the distance of inner edge from the rotational centre. Thus the farther edge of a spinning disc moves faster than the nearer edge.

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we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer