Question

Suppose a yo-yo has a center shaft that has a 0.250 cm radius and that its string is being pulled.

(a) If the string is stationary and the yo-yo accelerates away from it at a rate of 1.50 m/s2, what is the angular acceleration of the yo-yo?
(b) What is the angular velocity after 0.750 s if it starts from rest?
(c) The outside radius of the yo-yo is 3.50 cm. What is the tangential acceleration of a point on its edge? Ans:

(a) 600 rad/s2,
(b) 450 rad/s,
(c) 21.0 m/s

Answer 1

Answer:

600 rad/s²

450 rad/s

21 m/s²

Explanation:

r = Radius

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity

$\alpha$ = Angular acceleration

t = Time taken

Linear acceleration is given by

$a=r\alpha\\\Rightarrow \alpha=\dfrac{a}{r}\\\Rightarrow \alpha=\dfrac{1.5}{0.25\times 10^{-2}}\\\Rightarrow \alpha=600\ rad/s^2$

Angular acceleration of the yo-yo is 600 rad/s²

$\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=0+600\times 0.75\\\Rightarrow \omega_f=450\ rad/s$

angular velocity of the yo-yo is 450 rad/s

$a=r\alpha\\\Rightarrow a=3.5\times 10^{-2}\times 600\\\Rightarrow a=21\ m/s^2$

Tangential acceleration of a point on its edge is 21 m/s²

Answer 2

The angular acceleration, angular velocity and tangential acceleration are;

A) 600 rad/s²

B) 450 rad/s,

C) 21 m/s²

What is the angular velocity and acceleration?

A) We are given;

Radius; r = 0.25 cm = 0.0025 m

Linear acceleration; a = 1.5 m/s²

Formula for Angular Acceleration is;

α = a/r

Thus; α = 1.5/0.0025

α = 600 rad/s²

B) We are given;

time; t = 0.75 s

final angular velocity is gotten from;

ω_f = ω_i + αt

ω_f = 0 + (600 * 0.75)

ω_f = 450 rad/s

C) We are given;

Radius; r = 3.5 cm = 0.035 m

Formula for tangential acceleration is;

α_t = rα

α = 0.035 * 600

α = 21 m/s²

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