a compact disk, which has a diameter of $12.0\text{ cm}$, speeds up uniformly from zero to $4.50\text{ rev/s}$ in $3.05\text{ s}$.

part a
what is the tangential acceleration of a point on the outer rim of the disk at the moment when its angular speed is $2.00\text{ rev/s}$?
$a_t = $ $\text{m/s}^2$

part b
what is the tangential acceleration of a point on the outer rim of the disk at the moment when its angular speed is $3.00\text{ rev/s}$?
$a_t = $ $\text{m/s}^2$

Question

a compact disk, which has a diameter of $12.0\text{ cm}$, speeds up uniformly from zero to $4.50\text{ rev/s}$ in $3.05\text{ s}$.

part a
what is the tangential acceleration of a point on the outer rim of the disk at the moment when its angular speed is $2.00\text{ rev/s}$?
$a_t = $ $\text{m/s}^2$

part b
what is the tangential acceleration of a point on the outer rim of the disk at the moment when its angular speed is $3.00\text{ rev/s}$?
$a_t = $ $\text{m/s}^2$

Accepted Answer

{ "quality": "clear", "question_count": 2, "skills_matched": [ "step_cot", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Rotational Kinematics" ], "new_concepts": [ "Tangential Acceleration" ], "current_concepts": [ "Rotational Kinematics", "Tangential Acceleration" ] } ## Calculate the constant angular acceleration of the disk $$ \begin{aligned} d &= 12.0\text{ cm} \implies R = 6.00\text{ cm} = 0.0600\text{ m} \ \omega_0 &= 0\text{ rev/s} \ \omega_f &= 4.50\text{ rev/s} = 4.50 \times 2\pi\text{ rad/s} = 9\pi\text{ rad/s} \ t &= 3.05\text{ s} \ \alpha &= \frac{\omega_f - \omega_0}{t} = \frac{9\pi}{3.05}\text{ rad/s}^2 \approx 9.2704\text{ rad/s}^2 \end{aligned} $$ ## Determine the tangential acceleration for Part A $$ \begin{aligned} a_t &= R\alpha \ a_t &= 0.0600\text{ m} \times 9.2704\text{ rad/s}^2 \approx 0.556\text{ m/s}^2 \end{aligned} $$ ## Determine the tangential acceleration for Part B $$ \begin{aligned} a_t &= R\alpha \ a_t &= 0.0600\text{ m} \times 9.2704\text{ rad/s}^2 \approx 0.556\text{ m/s}^2 \end{aligned} $$ ## Question 1 The tangential acceleration of a point on the outer rim of the disk at the moment when its angular speed is $2.00\text{ rev/s}$ is 0.556 $\text{m/s}^2$. ## Question 2 The tangential acceleration of a point on the outer rim of the disk at the moment when its angular speed is $3.00\text{ rev/s}$ is 0.556 $\text{m/s}^2$. { "subject": "Natural Science", "question_type": "Multi-part", "knowledge_point": [ "Natural Science", "Physics", "Rotational Kinematics" ] }

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QUESTION IMAGE

Question

Explanation:

Calculate the constant angular acceleration of the disk

Determine the tangential acceleration for Part A

Determine the tangential acceleration for Part B

Answer:

Question 1

Question 2