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Question-212170

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Question Number 212170 by ajfour last updated on 04/Oct/24
Commented by ajfour last updated on 08/Oct/24
Commented by ajfour last updated on 08/Oct/24
I suspect it might happen this way.
$${I}\:{suspect}\:{it}\:{might}\:{happen}\:{this}\:{way}. \\ $$
Commented by mr W last updated on 08/Oct/24
i think too.
$${i}\:{think}\:{too}. \\ $$
Commented by ajfour last updated on 08/Oct/24
Answered by Spillover last updated on 06/Oct/24
torque about point a τ=mgrsinθ Angular acceration τ=I.α......(i) I=moment of initial of the ball I=(2/5)mr^2 (solid sphere)....(ii) τ=(2/5)mr^2 ×α α=(5/(2mr^2 ))τ from Angular motion θ(t)=θ_o +w_o t+(1/2)αt^2 .....(iii) θ(t)=θ_o +w_o t+(1/2)×(5/(2mr^2 ))τt^2 The angle turned can be calculated if we have specific value of r,m ,θ,w_o and t
$${torque}\:{about}\:{point}\:{a} \\ $$$$\tau={mgr}\mathrm{sin}\theta \\ $$$${Angular}\:{acceration} \\ $$$$\tau={I}.\alpha……\left({i}\right) \\ $$$${I}={moment}\:{of}\:{initial}\:{of}\:{the}\:{ball} \\ $$$${I}=\frac{\mathrm{2}}{\mathrm{5}}{mr}^{\mathrm{2}} \left({solid}\:{sphere}\right)….\left({ii}\right) \\ $$$$\tau=\frac{\mathrm{2}}{\mathrm{5}}{mr}^{\mathrm{2}} ×\alpha \\ $$$$\alpha=\frac{\mathrm{5}}{\mathrm{2}{mr}^{\mathrm{2}} }\tau \\ $$$${from}\:{Angular}\:{motion} \\ $$$$\theta\left({t}\right)=\theta_{{o}} +{w}_{{o}} {t}+\frac{\mathrm{1}}{\mathrm{2}}\alpha{t}^{\mathrm{2}} \:\:…..\left({iii}\right) \\ $$$$\theta\left({t}\right)=\theta_{{o}} +{w}_{{o}} {t}+\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{5}}{\mathrm{2}{mr}^{\mathrm{2}} }\tau{t}^{\mathrm{2}} \\ $$$${The}\:{angle}\:{turned}\:{can}\:{be}\:{calculated}\:{if} \\ $$$${we}\:{have}\:{specific}\:{value}\:{of}\:{r},{m}\:,\theta,{w}_{{o}} \:{and}\:{t} \\ $$
Commented by ajfour last updated on 06/Oct/24
Why does the ball turn about its center as well, i wished to know that..
$${Why}\:{does}\:{the}\:{ball}\:{turn}\:{about}\:{its}\:{center}\: \\ $$$${as}\:{well},\:{i}\:{wished}\:{to}\:{know}\:{that}.. \\ $$
Commented by Spillover last updated on 06/Oct/24
The ball turns about its center of mass due to the principles of rotational motion and the way forces act on it. 1 Center of Mass and Torque- The center of mass (COM) of a solid ball is at its geometric center. When the ball is released, it experiences gravitational force acting downward through its COM.- The torque that causes the ball to rotate is generated by the gravitational force acting at a distance from the pivot point (the attachment point of the arm). 2 Pivot Point and Rotation- As the ball swings, it rotates about the pivot point, but because it is a solid object, it also rotates about its own center of mass.- The ball's motion can be thought of as a combination of translational motion (the motion of the COM) and rotational motion (the spinning about its center). 3 Angular Motion- The ball can be modeled as a rigid body rotating about an axis (the pivot point) while simultaneously translating through space.- The angular motion of the ball about its COM and the pivot point is related through the parallel axis theorem, which accounts for the distribution of mass. Conservation of Angular Momentum- If no external torques act on the system, the angular momentum of the ball about its center of mass is conserved.- As it swings down, the ball's angular velocity changes due to gravitational acceleration, but it continues to rotate about its own center as well. 3 Visualizing the Motion- Imagine holding a ball on a string and letting it go. It swings down due to gravity while also spinning about its center. The same principle applies here, with the ball turning about both the pivot and its own center
$$ \\ $$The ball turns about its center of mass due to the principles of rotational motion and the way forces act on it.
1
Center of Mass and Torque- The center of mass (COM) of a solid ball is at its geometric center. When the ball is released, it experiences gravitational force acting downward through its COM.- The torque that causes the ball to rotate is generated by the gravitational force acting at a distance from the pivot point (the attachment point of the arm).
2
Pivot Point and Rotation- As the ball swings, it rotates about the pivot point, but because it is a solid object, it also rotates about its own center of mass.- The ball's motion can be thought of as a combination of translational motion (the motion of the COM) and rotational motion (the spinning about its center).
3
Angular Motion- The ball can be modeled as a rigid body rotating about an axis (the pivot point) while simultaneously translating through space.- The angular motion of the ball about its COM and the pivot point is related through the parallel axis theorem, which accounts for the distribution of mass.
Conservation of Angular Momentum- If no external torques act on the system, the angular momentum of the ball about its center of mass is conserved.- As it swings down, the ball's angular velocity changes due to gravitational acceleration, but it continues to rotate about its own center as well.
3
Visualizing the Motion- Imagine holding a ball on a string and letting it go. It swings down due to gravity while also spinning about its center. The same principle applies here, with the ball turning about both the pivot and its own center
$$ \\ $$
Answered by mr W last updated on 08/Oct/24
if the ball doesnt rotate, then the forces mg and T must pass through the center of ball, like this:
$${if}\:{the}\:{ball}\:{doesnt}\:{rotate},\:{then}\:{the} \\ $$$${forces}\:{mg}\:{and}\:{T}\:{must}\:{pass}\:{through} \\ $$$${the}\:{center}\:{of}\:{ball},\:{like}\:{this}: \\ $$
Commented by mr W last updated on 08/Oct/24
Commented by mr W last updated on 08/Oct/24
but in this case the ball has rotated an angle θ. this is contradiction. that means the ball must rotate. when the ball rotates, the force T should not pass through the center of ball. it must create a torque about the center of ball corresponding to the rotation angle, like this:
$${but}\:{in}\:{this}\:{case}\:{the}\:{ball}\:{has}\:{rotated} \\ $$$${an}\:{angle}\:\theta.\:{this}\:{is}\:{contradiction}. \\ $$$${that}\:{means}\:{the}\:{ball}\:{must}\:{rotate}. \\ $$$${when}\:{the}\:{ball}\:{rotates},\:{the}\:{force}\:{T} \\ $$$${should}\:{not}\:{pass}\:{through}\:{the}\:{center} \\ $$$${of}\:{ball}.\:{it}\:{must}\:{create}\:{a}\:{torque}\:{about} \\ $$$${the}\:{center}\:{of}\:{ball}\:{corresponding}\:{to} \\ $$$${the}\:{rotation}\:{angle},\:{like}\:{this}: \\ $$
Commented by mr W last updated on 08/Oct/24
Commented by mr W last updated on 08/Oct/24
rotation of ball: ϕ (clockwise) 0<ϕ<θ
$${rotation}\:{of}\:{ball}:\:\varphi\:\left({clockwise}\right) \\ $$$$\mathrm{0}<\varphi<\theta \\ $$
Commented by ajfour last updated on 08/Oct/24
y=asin θ+rsin φ x=acos θ+rcos φ ⇒ (y−asin θ)^2 +(x−acos θ)^2 =r^2 Torque about Origin(pivot) mgx=m((2/5)r^2 +x^2 +y^2 )ω_θ ((dω_θ /dθ)) mgy=(1/2)mv^2 +(1/2)(((2mr^2 )/5))ω_θ ^2 v^2 =((dx/dt))^2 +((dy/dt))^2 ...
$${y}={a}\mathrm{sin}\:\theta+{r}\mathrm{sin}\:\phi \\ $$$${x}={a}\mathrm{cos}\:\theta+{r}\mathrm{cos}\:\phi \\ $$$$\Rightarrow\:\:\left({y}−{a}\mathrm{sin}\:\theta\right)^{\mathrm{2}} +\left({x}−{a}\mathrm{cos}\:\theta\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${Torque}\:{about}\:{Origin}\left({pivot}\right) \\ $$$${mgx}={m}\left(\frac{\mathrm{2}}{\mathrm{5}}{r}^{\mathrm{2}} +{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\omega_{\theta} \left(\frac{{d}\omega_{\theta} }{{d}\theta}\right) \\ $$$${mgy}=\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{2}{mr}^{\mathrm{2}} }{\mathrm{5}}\right)\omega_{\theta} ^{\mathrm{2}} \\ $$$${v}^{\mathrm{2}} =\left(\frac{{dx}}{{dt}}\right)^{\mathrm{2}} +\left(\frac{{dy}}{{dt}}\right)^{\mathrm{2}} \\ $$$$… \\ $$

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