A wheel 1.60 m in diameter lies in a vertical plane and rotates about its centra

Question

A wheel 1.60 m in diameter lies in a vertical plane and rotates about its central axis with a constant angular acceleration of 3.55 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 angular speed of the wheel
rad/s

(b) the tangential speed of the point P
m/s

(c) the total acceleration of the point P

(d) the angular position of the point P
rad

Explanation / Answer

here,

diameter of the wheel , d = 1.6 m

radius , r = 0.8 m

angular accelration , a = 3.55 rad/s^2

(a)

at t = 2 s

let the angular speed be w

w = w0 + a*t

w = 0 + 3.55 * 2

w = 7.1 rad/s

the angular speed of the wheel is 7.1 rad/s

(b)

the tangential speed at point P , v = r * w

v = 0.8 * 7.1

v = 5.68 m/s

the tangential speed at point P is 5.68 m/s

(c)

centripital accelration , ac = v^2/r

ac = 5.68^2 /0.8

ac = 40.33 m/s^2

the total accelration , at = sqrt( ac^2 + a^2)

at = sqrt( 40.33^2 + 3.55^2)

at = 40.48 m/s^2

the total accelration is 40.48 m/s^2

theta = arctan(40.48/3.55)

theta = 84.99 degree

the direction is 84.99 degree to the radius of point P

(d)

distance covered , theta = 0 + 0.5 * a * t^2

theta = 0 + 0.5 * 3.55 * 2^2

theta = 7.1 rad

theta = 407 degree

the total angle , theta' = theta + 57.3 degree

theta' = 464.31 degree

theta' = 8.09 rad

and the angular position of point P is 1.82 rad from the horizontal

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