Question

A uniform disk with radius 0.650 m

Answer 1

Answer:

$a = 13.758\,\frac{m}{s^{2}}$

Explanation:

First, the instant associated to the angular displacement is:

$(1.10\,\frac{rad}{s} )\cdot t + (6.30\,\frac{rad}{s^{3}} )\cdot t^{2} - 0.628\,rad = 0$

Roots of the second-order polynomial are:

$t_{1} \approx 0.240\,s, t_{2} \approx -0.415\,s$

Only the first root is physically reasonable.

The angular velocity is obtained by deriving the angular displacement function:

$\omega (0.240\,s) = 1.10\,\frac{rad}{s}+ (12.6\,\frac{rad}{s^{2}})\cdot (0.240\,s)$

$\omega (0.240\,s) = 4.124\,\frac{rad}{s}$

The angular acceleration is obtained by deriving the previous function:

$\alpha (0.240\,s) = 12.6\,\frac{rad}{s^{2}}$

The resultant linear acceleration on the rim of the disk is:

$a_{t} = (0.650\,m)\cdot (12.6\,\frac{rad}{s^{2}} )$

$a_{t} = 8.190\,\frac{m}{s^{2}}$

$a_{n} = (0.650\,m)\cdot (4.124\,\frac{rad}{s} )^{2}$

$a_{n} = 11.055\,\frac{m}{s^{2}}$

$a = \sqrt{a_{t}^{2}+a_{n}^{2}}$

$a = 13.758\,\frac{m}{s^{2}}$