Question

a particle is moving along a circular path having a radius of 4 in such that its position as a function of time is given by theta = cos 2t, where theta is in radians and t is in seconds. Determine the magnitude of the accer

Answer 1

Step-by-step explanation:

Given

radius of circular path
`r=4\ in.`

Position is given by

`\theta =\cos 2t---1`

Differentiate 1 to angular velocity we get

`\frac{\mathrm{d} \theta }{\mathrm{d} t}=\omega =-2\sin 2t----2`

Differentiate 2 to get angular acceleration

`\frac{\mathrm{d} \omega }{\mathrm{d} t}=-2^2\cos 2t ---3`

Net acceleration is the vector summation of tangential and centripetal force

`a_t=\alpha * r`

`a_t=-4\cos 2t* 4=-16\cos 2t`

`a_r=\omega ^2\cdot r`

`a_r=(-2\sin 2t)^2\cdot 4`

`a_r=16\sin^2(2t)`

`a_(net)=√(a_r^2+a_t^2)`

`a_(net)=√((16\sin ^2(2t)+(-16\cos 2t)^2)`

`a_(net)=√(256\cos ^2(2t)+256\sin ^4(2t))`