# Question on Moment of Inertia/Rotational Inertia

## Homework Statement

In the figure, a wheel of radius 0.42 m is mounted on a frictionless horizontal axle. A massless cord is wrapped around the wheel and attached to a 2.7 kg box that slides on a frictionless surface inclined at angle θ = 28 owith the horizontal. The box accelerates down the surface at 1.9 m/s2. What is the rotational inertia of the wheel about the axle?

Diagram attached

## Homework Equations

Torque=Moment of Inertia*angular acceleration
F=ma

## The Attempt at a Solution

what i did was combinr the equations and yielded the following
I=(F*r^2)/a=0.47628
I don't see where I did wrong but the solution does not match

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haruspex
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I=(F*r^2)/a
You don't say how you calculate the tension. We cannot tell where you went wrong if you do not post your working.

I just realised what I calculated was Fa instead of tension. But how do i find the tension?? I dont know what to do with the angle ø=28°. pls help
You don't say how you calculate the tension. We cannot tell where you went wrong if you do not post your working.

haruspex
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