Now let\'s consider the tangential and radial components of a discus as it is be

Question

Now let's consider the tangential and radial components of a discus as it is being thrown. A discus thrower turns with angular acceleration =50rad/s2, moving the discus in a circle of radius 0.80 m (Figure 1) . Find the radial and tangential components of acceleration of the discus (modeled as a point) and the magnitude of its acceleration at the instant when the angular velocity is 10 rad/s .

SET UP As shown in (Figure 2) , we model the thrower's arm as a rigid body, so r is constant (this may not be completely realistic). We model the discus as a particle moving in a circular path.

SOLVE The components of the acceleration are given by

arad = ^2r=(10rad/s)^2(0.80m)=80m/s^2

atan = r=(0.80m)(50rad/s^2)=40m/s^2

The magnitude of the acceleration vector is

a= (square root of) arad^2 + atan^2

=89m/s^2

REFLECT The magnitude of the acceleration is about nine times the acceleration due to gravity; the corresponding force supplied by the thrower's arm must be about nine times the weight of the discus. Note that we have omitted the unit radian in our final results; we can do this because a radian is a dimensionless quantity.

Part A: What is the direction of the acceleration vector of the discus if the angular acceleration is 33 rad/s2 and =8.0rad/s?

Express your answer in degrees to two significant figures. Measure the angle from the radial line between the center of the discus and the thrower's shoulder.     

Explanation / Answer

alpha = 50 rad/s^2

r = 0.80 m

tanential acc = alpha r = 40 m/s^2 ......Ans

radial acc = w^2 r = 80 m/s^2 ......Ans

magnitude of acc. = sqrt(a_r^2 + a_t^2)

= 89.4 m/s^2

-------------

a_r = 8^2 (r) = 64 r

a_t = 33 r

theta = tan^-1 [ a_t / a_r]

= 27.3 deg .....Ans

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