The wheel of radius r 300 mm rolls to the right without sli

The wheel of radius r = 300 mm rolls to the right without slipping and has a velocity v_0 = 3m/s and an acceleration a_0 =-5m/s^2of its center 0 as shown in the figure. Calculate the velocity of point A on the wheel for the instant represented. Calculate the acceleration of point A on the wheel for the instant represented.

Solution

a)

w = V_O / r = 3/0.3 = 10 rad/s

V_A = V_O + V_A/O

V_O = 3i

V_A/O = w X r_A/O

= 10 (-k) X 0.2(-i cos30 + j sin30)

= -10k X (-0.1732i + 0.1j)

= i + 1.732j

v_A = 3i + (i + 1.732j)

v_A = 4i + 1.732j

Magnitude of v_A = sqrt (4² + 1.732²) = 4.36 m/s

b)

Angular acceleration, alpha = a_O /r = -5 / 0.3 = -16.67 rad/s²

In vector notation, alpha = 16.67k rad/s²

a_A = a_O + a_A/O

a_O = -5i

a_A/O = w X (w X r) + (alpha X r)

= (-10k) X (i + 1.732j) + (16.67k X 0.2(-i cos30 + j sin30))

= (17.32i -10j) + (16.67k X (-0.1732i + 0.1j))

= (17.32i -10j) + (-1.6667i - 2.8867j)

= 15.653i - 12.8867j

a_A = -5i + (15.653i - 12.8867j)

a_A = 10.653i - 12.8867j

Magnitude = sqrt (10.653² + 12.8867²) = 16.72 m/s²