The problem is as follows:

A pulley starts spinning from rest a rotation with constant angular acceleration. After $5\,s$ a point in its periphery has an instant acceleration which makes a $53^{\circ}$ angle with its linear speed. Find the modulus of the angular acceleration (in $\frac{rad}{s^{2}}$) of the pulley.

The given alternatives in my book are as follows:

$\begin{array}{ll} 1.&0.5\,\frac{m}{s}\\ 2.&0.53\,\frac{m}{s}\\ 3.&0.053\,\frac{m}{s}\\ 4.&0.106\,\frac{m}{s}\\ 5.&1.06\,\frac{m}{s}\\ \end{array}$

For this particular problem. I'm lost as how should I use the given information of the instant acceleration and the linear speed. How should I put those vectors?. Which sort of equation should I use?.

The only equation which comes to my mind for the angular acceleration is how it is related to the tangential acceleration as:

$a_{t}=\alpha \times r$

But in this case there is no radius.

Thus I believe it has something to do with vectors but I can't really find exactly how to use that information. Can somebody help me here?.

  • $\begingroup$ Is $v(5)$ or the magnitude of the acceleration given? $\endgroup$ Nov 4, 2019 at 11:13
  • $\begingroup$ @Certainly not a dog No, it is not given. But it looks that is unnecesary. Perhaps does it mean that cannot be solved? $\endgroup$ Nov 4, 2019 at 17:22

1 Answer 1


let distance of point from centre of pulley be r let angular acc. = a write tangential acc. = r*a...........(1) write normal acc. = (w^2)*r.......(2) where w is angular velocity also w at t= 5 seconds is a*5 put w in (2) also tan(53)=normal acc./tangential acc. (this is because instantaneous linear velocity is along tangent at point) you will see that r is cancelled and a= 0.053

  • $\begingroup$ The part of instantaneus acceleration making an angle with that of linear speed is where I am stuck at. In the passage it mentions this. But you seem not using that information and just jumping to tangential acceleration and centripetal acceleration. The modulus of the centripetal acceleration doesn't affect the angle of the instantaneus acceleration the question is referring to? Can you clarify this please? $\endgroup$ Nov 4, 2019 at 17:28
  • $\begingroup$ This answer is very difficult to read because of the poor formatting. Hint: to start a new line in the displayed text you need either a blank line in the editing window or you need two blank characters at the end of the line in the editing window. (I recommend a blank line.) In order to make the formulas more readable, use MathJax: math.stackexchange.com/help/notation $\endgroup$
    – David K
    Nov 4, 2019 at 18:25
  • $\begingroup$ Despite the poor formatting this is a correct answer. The instant acceleration that makes a 53-degree angle in the question is the vector sum of tangential and centripetal acceleration, so certainly the modulus of centripetal acceleration does affect the angle and this answer does not say it doesn't. The answer says that when you correctly account for both tangential and centripetal acceleration, the factor of $r$ that is in both of those accelerations will drop out. $\endgroup$
    – David K
    Nov 4, 2019 at 18:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.