Very basic question on particle dynamics

[TomW17]

Homework Statement

A car makes a left turn from a stopped position, increasing its speed at a rate of 2 m/s/s. If a book is on the dashboard of the car, at what time will the book begin to slide if the static coefficient of friction \mu_s = 0.3? The radius of the curve of the motion is 7m.

Seems straight forward enough but the answer I get is different to the one my teacher put in the tutorial sheet, and would just like a 2nd opinion on this to make sure I'm not going mad.

Homework Equations

Normal force on a particle in circular motion: F_n = m\frac{v^2}{R}
Maximum frictional force between the box and dash board: F_f = N\mu_s = mg \mu_s
Constant acceleration equations: v = v_0 + at

The Attempt at a Solution

Let the book have mass m. Just as the book is about to slide: F_f = F_n \implies m\frac{v^2}{R} = m\frac{(at)^2}{R} = mg \mu_s \implies t = \frac{\sqrt{gR\mu_s}}{a}

Plugging in the numbers, I get t\approx 2.27 \text{s} whereas my teacher's put t= 1.94\text{s}

Thanks for the help.

 

[collinsmark]

Don't forget there are two components of the acceleration. There's the radial (centripetal) acceleration that you have calculated above in your original post.

But there is also the tangential acceleration (2 m/s²) that acts at a right angle relative to the radial acceleration.

You can find the total acceleration's magnitude by finding the hypotenuse.

 

[TomW17]

Don't forget there are two components of the acceleration. There's the radial (centripetal) acceleration that you have calculated above in your original post.

But there is also the tangential acceleration (2 m/s²) that acts at a right angle to the radial acceleration.

You can find the total acceleration's magnitude by finding the hypotenuse.

But surely the centripetal acceleration has no effect on the magnitude of the book's tangential velocity (speed)? So the constant acceleration equation still applies to calculate the book's speed, which in turn can be used to calculate the magnitude of the centripetal force, which the frictional force between the book and the dashboard needs to match to prevent it from sliding, no?

 

[collinsmark]

But surely the centripetal acceleration has no effect on the magnitude of the book's tangential velocity (speed)? So the constant acceleration equation still applies to calculate the particle's speed, which in turn can be used to calculate the magnitude of the centripetal force?

Yes, go ahead and use your |v| = |a_t|t equation to find the book's speed (that's the way it was specified in the problem statement, in less specific variables). And like you've already done, you can use a_r = \frac{v^2}{R}, as you have done.

But then note that the overall acceleration magnitude is the magnitude of the two acceleration vectors forming a right triangle.

 

[TomW17]

Yes, go ahead and use your |v| = |a_t|t equation to find the book's speed (that's the way it was specified in the problem statement, in less specific variables). And like you've already done, you can use a_r = \frac{v^2}{R}, as you have done.

But then note that the overall acceleration magnitude is the magnitude of the two acceleration vectors forming a right triangle.

Aha, got it, completely neglected that the frictional force accelerates the book tangentially as well as keeps it in circular motion. Thanks very much.