Question

A disk is uniformly accelerated from rest with angular acceleration α. The magnitude of the linear acceleration of a point on the rim of the disk (a) grows with the time t as ________ for (αt2) << 1.

Answer 1

The linear acceleration of a point on the rim of the disk grows linearly with time, represented by the equation a = αr, where α is the angular acceleration and r is the disk radius. This linear relationship holds when the product of angular acceleration and time squared (αt2) is much smaller than 1 and the radius is constant.

Step-by-step explanation:

The student's question asks how the magnitude of the linear acceleration of a point on the rim of a disk grows with time if the disk experiences uniform angular acceleration from rest and (αt2) << 1. Given that the tangential acceleration (a) is the product of angular acceleration (α) and the radius of the disk (r), for small angles, the linear acceleration grows linearly with time, assuming a constant radius. The formula for this is a = αr.

In the provided context, assuming small-time intervals where (αt2) is much less than 1, and the angular acceleration is constant, the linear acceleration a at the rim of the disk increases proportionally to the angular acceleration α and the radius r, which is not changing with time. This implies that for small durations after the disk starts from rest, the linear acceleration can be approximated as a linear function of time t provided the angular acceleration is also linear with time and the condition (αt2) << 1 holds.

Answer 2

`a = R\alpha√(1 + \alpha^2t^4)`

Step-by-step explanation:

As we know that the acceleration of a point on the rim of the disc is in two directions

1) tangential acceleration which is given as

`a_t = R\alpha`

2) Centripetal acceleration

`a_c = \omega^2 R`

here we know that

`\omega = \alpha t`

`a_c = (\alpha t)^2 R`

now we know that net linear acceleration is given as

`a = √(a_c^2 + a_t^2)`

so we have

`a = √(R^2\alpha^2 + R^2(\alpha t)^4)`

`a = R\alpha√(1 + \alpha^2t^4)`