A wheel 1.55 m in diameter lies in a vertical plane and rotates about its centra

Question

A wheel 1.55 m in diameter lies in a vertical plane and rotates about its central axis with a constant angular acceleration of 4.20 rad/s2. The wheel starts at rest at t = 0, and the radius vector of a certain point P on the rim makes an angle of 57.3° with the horizontal at this time. At t = 2.00 s, find the following.

a) the total acceleration of the point P

magnitude= 55.32

direction in degrees with respect to the radius to point P?

b) (d) the angular position of the point P in radians

magnitude= 55.32

direction in degrees with respect to the radius to point P?

b) (d) the angular position of the point P in radians

Explanation / Answer

here

w = 2 * 4.2 = 8.4 rad/s

then

v = r * w

v = (1.55/2) * 8.4 = 6.51 m/s

for tengential acceleration

at = r * alpha

at = (1.55/2) * 4.2 = 3.255 m/s^2

then

ar = v^2 / r = 6.51^2 / (1.55 / 2) = 54.684 m/s^2

magnitude of acceleration is = sqrt( at^2 + ar^2)

a = sqrt( 3.255^2 + 54.684^2)

a = 54.8 m/s^2

then the direction is

tan(theta) = at/ar = 3.255 / 54.684

theta = 3.406 deg

d)

by using the second equation of motion

angular position = 57.3 * pie/180 + 0.5 * 4.2 * 2^2

= 9.4 rad

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