+10 A basketball player is spinning a basketball, of radius 23 cm and mass 0.625 kg, on the tip of his finger. If he accelerates the ball at 0.3 m/s2, what torque is he applying on the ball? Assume his finger is frictionless. How do you do this? I keep getting the wrong answer of .007
+37093 F= ma and torque = f x d where d= the radius of 0.23 m
= ma x r
=.625 (.3)(.23) = .0431 N-m
+10 Furthermore, I left out that they provided \(I=2/3mr^2\) . is that extraneous information? bc i recall that \(T=I\alpha \)
+37093 " When a torque is applied to an object it begins to rotate with an acceleration inversely proportional to its moment of inertia."
SO its moment of inertia ( 2/3 mr^2 [for a thin walled sphere]) would have been used to determine the acceleration...which was already computed (if you will) and given to you in the question as 0.3 m/s^2....so ...yah, I guess it is not needed unless you used it to calculate the acceleration you gave in your question.
Please let me know if you find something here to be in error....Thanx !
+37093 RE-thinking this a bit....
T = a * (2/3 mr^2) = .3 * (2/3 (.625)(.23^2)) = .0066125 N-m BUT I do not think this is correct either as your question gives acceleration as .3 m/s^2 which is a LINEAR acceleration......so:
Are we to calculate the angular acceleration from THAT or were you actually given ANGULAR acceleration ????
Let me know and I will continue re-learning this stuff ! (it has been a while)....keep me posted....
+33653 angular acceleration = linear acceleration/radius, so here: \(\alpha = 0.3/0.23s^{-2}\)
\(T=\frac{2}{3}\times 0.625\times 0.23^2\times \frac{0.3}{0.23} Nm\)
+37093 Thanx Alan....that is what I was (eventually) thinking....but was unsure of the question. Cool...
=.0727 N-m
