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You have your bicycle upside down for repairs, with is 66.0 cm diameter wheel spinning freely at 230 rpm. The wheel's mass is 1.90 kg, concentrated mostly at the rim. You hold a wrench against the tire for 3.10 s, applying a 2.70 N normal force. If the coefficient of friction between wrench and tire is 0.460, what is the final angular speed of the wheel?

Respuesta :

Answer:

[tex]\omega_f = 17.86\ rad/s[/tex]

Explanation:

given,

dia of wheel = 66 cm

radius = 33 cm = 0.33 m

mass of wheel = 1.90 Kg

speed of wheel = 230 rpm

time to hold the tire  = 3.10 s

Normal force = 2.7 N

coefficient of friction = 0.46

final angular speed = ?

we know

F = μ N

F = 0.46 x 2.7 = 1.242 N

torque

τ = F r

τ = 1.242 x 0.33 = 0.409 Nm

moment of inertia for wheel

I = MR²

I = 1.90 x 0.33²

I = 0.2069 kg m²

we now,

τ = I x α

[tex]\alpha = \dfrac{\tau}{I}[/tex]

[tex]\alpha = \dfrac{0.409}{0.2069}[/tex]

α = 1.981 rad/s²

we know,

[tex]\omega_f - \omega_i = \alpha\ t[/tex]

[tex]\omega_f - (230\dfrac{2\pi}{60}) =- 1.981 \times 3.1[/tex]

[tex]\omega_f - 24 = - 6.14[/tex]

[tex]\omega_f = 17.86\ rad/s[/tex]