Physics, 14.04.2020 18:26 ethangorrell67

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?

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Answer from: mollie053005

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

Answer from: Quest

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2

Answer from: Quest

there are no components of a physical goal mentioned on the list of choices that you've provided.

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