A 2.3kg bicycle wheel has a diameter of 50cm. What torque must you apply to take the wheel from 0rpm to 120rpm in 5.5s?

τ = 0.26 N.m

Explanation:

First we find the moment of inertia of the wheel, by using the following formula:

I= mr²

where,

I = Moment of Inertia = ?

m = mass of wheel = 2.3 kg

r = radius of wheel = 50 cm/2 = 25 cm = 0.25 m

Therefore,

I = (2.3 kg)(0.25 m)²

I = 0.115 kg.m²

Now, we find the angular acceleration of the wheel:

α = (ωf - ωi)/t

where,

α = angular acceleration = ?

ωf = final angular velocity = (120 rpm)(2π rad/1 rev)(1 m/60 s) = 12.56 rad/s

ωi = Initial Angular Velocity = 0 rad/s

t = time = 5.5 s

Therefore,

α = (12.56 rad/s - 0 rad/s)/(5.5 s)

α = 2.28 rad/s²

Now, the torque is given as:

Torque = τ = Iα

τ = (0.115 kg.m²)(2.28 rad/s²)

τ = 0.26 N.m

a displacement distance is size of displ

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer