Question

uniform disk with mass 40.0 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is stationary. The disk is initially at rest, and then a constant force 퐹=30.0푁is applied tangent to the rim of the disk.a) What is the magnitude 푣of the tangential velocity of a point on the rim of the disk after the disk has turned through 0.200 revolution?b) What is the magnitude 푎of the resultant acceleration of that point in part a?

Answer 1

Answer:

The magnitude of the tangential velocity is $v= 0.868 m/s$

The magnitude of the resultant acceleration at that point is  $a = 4.057 m/s^2$

Explanation:

From the question we are told that

      The mass of the uniform disk is $m_d = 40.0kg$

       The radius of the uniform disk is $R_d = 0.200m$

       The force applied on the disk is $F_d = 30.0N$

Generally the angular speed i mathematically represented as

             $w = \sqrt{2 \alpha \theta}$

Where $\theta$ is the angular displacement given from the question as

           $\theta = 0.2000 rev = 0.2000 rev * \frac{2 \pi \ rad }{1 rev}$

                 $=1.257\ rad$

   $\alpha$ is the angular acceleration which is mathematically represented as

                    $\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

    The moment of inertial is mathematically represented as

                     $I = \frac{1}{2} m_dR^2_d$

Substituting values

                    $I = 0.5 * 40 * 0.200^2$

                        $= 0.8kg \cdot m^2$

Considering the equation for angular acceleration

               $\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

Substituting values

               $\alph$$\alpha = \frac{(30.0)(0.200)}{0.8}$

                   $= 7.5 rad/s^2$

Considering the equation for angular velocity

    $w = \sqrt{2 \alpha \theta}$

Substituting values

     $w =\sqrt{2 * (7.5) * 1.257}$

         $= 4.34 \ rad/s$

The tangential velocity of a given point on the rim is mathematically represented as

                 $v = R_d w$

Substituting values

                    $= (0.200)(4.34)$

                     $v= 0.868 m/s$

The radial acceleration at hat point  is mathematically represented as

            $\alpha_r = \frac{v^2}{R}$

                  $= \frac{0.868^2}{0.200^2}$

                 $= 3.7699 \ m/s^2$

The tangential acceleration at that point is mathematically represented as

               $\alpha _t = R \alpha$

Substituting values

           $\alpha _t = (0.200) (7.5)$

                 $= 1.5 m/s^2$

The magnitude of resultant acceleration at that point is

                 $a = \sqrt{\alpha_r ^2+ \alpha_t^2 }$

Substituting values

                $a = \sqrt{(3.7699)^2 + (1.5)^2}$

                   $a = 4.057 m/s^2$