- #1
ubiquinone
- 43
- 0
Hi I've been doing more problems involving rotational motion. This question involves moments of inertia and torque which I'm not sure on how to solve. If anyone could help, that would be great. Thanks!
Question: The uniform 4.0 kg thin plate is made to rotate about a vertical axis that is perpendicular to the page and through the point indicated on the lower right corner. Forces are applied to the plate as indicated.
Calculate a) the moment of inertia about the plate about the indicated axis, b) the total torque about the axis through the corner, c) the angular acceleration of the plate and d) the magnitude of the initial acceleration of the center of mass on the plate when the mass starts from rest.
Part a) I think you use the parallel-axis theorem:
\displaystyle I = I_{cm}+Md^2
where \displaystyle I_{cm}=\frac{1}{12}M(c^2+b^2}=\frac{1}{12}(4.0kg)(.55^2+.25^2)m^2=0.122 kg\cdot m^2
\displaystyle d^2=\left(\frac{c}{2}\right)^2+\left(\frac{b}{2}\right)^2=\left(\frac{0.55}{2}\right)^2+\left(\frac{0.25}{2}\right)^2=0.0912m
Therefore, \displaystyle I=(0.122kg\cdot m^2) + (4.0kg)(0.0912m^2)=0.487kg\cdot m^2
Part b) Do you just find the individual moments and sum them up?
If so, \displaystyle \tau_1 = r_1\times F_1=+(0.25)(20)\sin 30^o = 2.5 N\cdot m, \displaystyle \tau_2 = r_2\times F_2 = -(\sqrt{0.55^2+0.25^2})(25)\sin(180^0-\tan^{-1}(\frac{0.25}{0.55}))= -6.25 N\dot m
and \displaystyle \tau_3 =r_3\times F_3 = +(0.55m)(8N)\sin 20^o =1.50 N\cdot m
Total torque = (2.5-6.25+1.50)N\cdot m = -2.25 N\cdot m
Part c) The angular acceleration of the plate, well can't we just use Newton's second law for rotational motion, \displaystyle \tau_{total}=I\alpha
Then, \displaystyle \alpha=\frac{\tau_{total}}{I}=\frac{-2.25 N\cdot m}{0.487kg\cdot m^2}=-4.62 rad/s^2
Part d) The magnitude of the initial acceleration of the center of mass on the plate when mass starts from rest. This could be found just by \displaystyle a=r\alpha where r=d?
If so, \displaystyle a=(-4.62rad/s^2)\sqrt{\left(\frac{0.55}{2}\right)^2+\left(\frac{0.25}{2}\right)^2}=-1.395m/s^2
Question: The uniform 4.0 kg thin plate is made to rotate about a vertical axis that is perpendicular to the page and through the point indicated on the lower right corner. Forces are applied to the plate as indicated.
Code:
Diagram:
F_2
/ F_1
/ \
/__________\
| |
| |
| |
b | x |
| |
| |
|___________|
/ c Axis
/F_3
x - marks the center of mass
F_1 = 20.0 N at 30 degrees
F_2 = 25.0 N at 40 degrees
F_3 = 8.0 N at 70 degrees
c = 55.0 cm
b = 25.0 cmPart a) I think you use the parallel-axis theorem:
\displaystyle I = I_{cm}+Md^2
where \displaystyle I_{cm}=\frac{1}{12}M(c^2+b^2}=\frac{1}{12}(4.0kg)(.55^2+.25^2)m^2=0.122 kg\cdot m^2
\displaystyle d^2=\left(\frac{c}{2}\right)^2+\left(\frac{b}{2}\right)^2=\left(\frac{0.55}{2}\right)^2+\left(\frac{0.25}{2}\right)^2=0.0912m
Therefore, \displaystyle I=(0.122kg\cdot m^2) + (4.0kg)(0.0912m^2)=0.487kg\cdot m^2
Part b) Do you just find the individual moments and sum them up?
If so, \displaystyle \tau_1 = r_1\times F_1=+(0.25)(20)\sin 30^o = 2.5 N\cdot m, \displaystyle \tau_2 = r_2\times F_2 = -(\sqrt{0.55^2+0.25^2})(25)\sin(180^0-\tan^{-1}(\frac{0.25}{0.55}))= -6.25 N\dot m
and \displaystyle \tau_3 =r_3\times F_3 = +(0.55m)(8N)\sin 20^o =1.50 N\cdot m
Total torque = (2.5-6.25+1.50)N\cdot m = -2.25 N\cdot m
Part c) The angular acceleration of the plate, well can't we just use Newton's second law for rotational motion, \displaystyle \tau_{total}=I\alpha
Then, \displaystyle \alpha=\frac{\tau_{total}}{I}=\frac{-2.25 N\cdot m}{0.487kg\cdot m^2}=-4.62 rad/s^2
Part d) The magnitude of the initial acceleration of the center of mass on the plate when mass starts from rest. This could be found just by \displaystyle a=r\alpha where r=d?
If so, \displaystyle a=(-4.62rad/s^2)\sqrt{\left(\frac{0.55}{2}\right)^2+\left(\frac{0.25}{2}\right)^2}=-1.395m/s^2
Last edited: