Recent content by chronorec

I don't really get what you mean by geometrical constraints. My understanding of the question is such that the string is wound tightly around the shaft so that slipping will not occur. If the end of the string is not connected to the shaft itself, then I think it may be possible that slipping...

Try using back the equation R=(u²sin(40)/g), notice that the maximum distance is directly proportion to u2, hence when the velocity is changed to 1.06 times, R will become 1.062 times.

I'm assuming it's pure rotational motion with no slipping, hence a=r(alpha) however I have doubts on the ascending part, is the acceleration of the yoyo really upwards or downwards? I'm taken it to be upwards in my calculations

Homework Statement A yo-yo is made from two uniform disks of raduis R with a combined mass of m. A short massless shaft of radius r connects the disks. A long thin string is wrapped around the shaft several times by a yo-yo player, who releases it with zero speed. Assuming that the string is...

I'm using S to denote the displacement of the object and u to denote the initial velocity of the object, g is the acceleration due to gravity(ie. 9.81) and R is the maximum range which is 7.8 is this case. Using 3.6tan(20) to denote the maximum height is wrong as you are assuming that the...

I'm not sure how you got this equation - 3.6tan(20)=1.42 but it's wrong The max height is given by Sy=(u2sin220)/2g, this is derived from the equation Vy² = Vy0² - 2gSy and letting Vy be zero. For maximum range you should use the equation, R=(u2sin(40)/g), this is derived using S=V0t+0.5at2...