A wheel of radius 6 m has center at 10 in above the ground. Take the vertical li

Question

A wheel of radius 6 m has center at 10 in above the ground. Take the vertical line through the center as the y-axis. and the ground as the r-axis. An object on the circumference is rotating counterclockwise at constant speed. It starts at (6, 10) at time t = 0 and makes 1 revolution every 3 seconds. After 1 second the object flies off the wheel and continues to travel at the same speed in the direction of the tangent line. We are neglecting air resistance and gravity in this problem. (a) Write parametric formulas for the object's position for 0 lessthanorequalto t lessthanorequalto 1. (b) What is the object's location when t = 1 sec, and what are its horizontal velocity and vertical velocity at that instant? (c) Write parametric formulas for the object's position for t greaterthanorequalto 1.

Explanation / Answer

a)

(x,y,) co-rodinates of a circle can usually be represented as (r cos a, r sin a)
In this case, since the starting position is (6,10) and the time period for 2pie distnace is 3 seconds;
x will become (6cos 2tpie/3) since time period of 3 seconds for 2 pie so for t seconds it will be 2pie/3
y wil become 10+ 6 sin 2tpie/3
thus as long as the particle is in circular motion the (x,y) co-ordinates are =
(6 cos 2tpie/3 , 10+ 6sin 2tpie/3) for 0<t<1;

b)

At t=1 second; x= 6cos2pie/3 = 6*(-1/2) = -3
y= 10 + 6 sin 2pie/3 = 10+3sqrt3
Thus at t=1 the location is (-3, 10+3sqrt(3) ) -> this is the location at which tangent is drawn to the circle
vx = dx/dt = -6 sin 2tpie/3 * 2pie/3 = -4pie sin 2tpie/3 ; at t=1 the vx = -4pie* sqrt3 / 2 = -2piesqrt(3)
vy = dy/dt = 6 cos 2tpie/3 * 2pie/3= 4 pie cost 2tpie/3; at t=1 the vy = 4pie* (-1/2) = -2pie
Thus horizontal velocity = -2pie sqrt (3) ; and vertical velocity = -2pie ;
Note: negative values of velocity means the object is heading in negative direction with respect to the axis;

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