- #1

Extremist223

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## Homework Statement

The wheel of fortune wheel, whose radius is 3.00m, has 20 different numbers, at equal intervals, for a contestant to land on. The second contestant can spin the wheel, and wishes to land on the $1.00. The wheel starts off 4 positions away from the $1.00. Assuming a deceleration of 0.400 rad/s2, what must be the minimum tangential velocity for the contestant to reach the $1.00, if they pull the wheel down?

## Homework Equations

2pi rads is a full circle

w^2 = wo^2 + 2(alpha)(theta)

alpha= angular acceleration

theta is distance in radians travelled

w^2 is final angular velocity squared

wo^2 is initial angular velocity squared

Tangential Velocity = Radius x w

## The Attempt at a Solution

2pi/20= 0.314 rads per interval

the goal is 4 intervals away from the start therefore 4x0.314= 1.257 rads travelled.

it wants to land 4 intervals away from the start so the final angular velocity = 0 rad/s

therefore w^2 = wo^2 + 2 (alpha)(theta)

0 = wo^2 + 2(-.4rad/s^2)(1.257)

-wo^2= -1.0056rad/s

wo^2 = 1.0056rad/s

wo= 1.00265rad/s

so at this point i took vt= rw vt= 3m x 1.00265rad/s vt = 3.008m/s

I'm getting the answer wrong and I don't know why. Can anyone help me please?