Problem 1
A crate of mass \(m\) and dimensions \(b\times b\times h\) is placed on a horizontal conveyor belt. It is at rest when the belt suddenly starts up at constant acceleration \(a\). The crate is modeled as a uniform rectangular cuboid. The contact is characterized by the static and kinetic coefficients of friction \(\mu_s\) and \(\mu_k\).
(Note: the body’s mass moment of inertia about an axis \(Gz\) normal to the plane of motion is \(I_{Gzz}= m \frac{b^2+h^2}{12}\)).
Answers
A. The crate does not slip nor tip as long as \(a < \min(\mu_s g, \frac{b}{h} g)\).
B. Assuming that the crate does not slip, the crate will not tip as long as $a < g $.
C. If the crate tips without slipping, its initial angular acceleration is given by \(\alpha_0 = \frac{3}{2(h^2+b^2)} (h a - b g)\) (in the counterclockwise direction).
D. If the crate tips without slipping, its initial angular acceleration is given by \(\alpha_0 = \frac{6}{(h^2+b^2)} (h a - b g)\) (in the counterclockwise direction).
E. The body cannot possibly both slip and tip.
Solution
To figure out which given answers are correct, the problem requires a complete solution.
Step 1: We define a basis of unit vectors \((\boldsymbol{\hat{\imath}},\boldsymbol{\hat{\jmath}},\boldsymbol{\hat{k}})\) as shown in the Figure. In the first of part of the analysis we assume that the body is in translation at acceleration \(a\boldsymbol{\hat{\imath}}\), then its angular velocity and acceleration vanish.
Step 2: We sketch a FBD to show the forces acting on the crate:
gravitational forces with resultant \((-mg \boldsymbol{\hat{\jmath}})\) applied to mass center \(G\),
the contact forces exerted by the belt on \({\cal B}\) whose resultant must lie in the plane of motion of \(G\) at some unknown point \(I\). The position of \(I\) is defined by distance \(x\). We write the resultant as \({\bf R}= N\boldsymbol{\hat{\jmath}}+ F \boldsymbol{\hat{\imath}}\) where \(N\) is the normal reaction and \(F\) the friction force.
Step 3: We apply Euler’s first principle \(m \boldsymbol{a}_G = {\bf F}_{ext\to {\cal B}} = F\boldsymbol{\hat{\imath}}+ (N-mg) \boldsymbol{\hat{\jmath}}\) and \({\bf M}_G = \boldsymbol{0}\) to get 3 equations \[ m a= F, \qquad N-mg = 0, \qquad \frac{h}{2} F -x N =0 \qquad{[1-3]} \] We also need to impose the conditions: \(|F|< \mu_s N\) to guarantee that the crate moves without slipping. We also must impose \(x < b/2\) to guarantee that the point \(I\) of application of the contact forces does not fall beyond the cornerpoint \(A\). With [1-3], we find \(F= ma\), \(N=mg\) and \(x= h a/2g\) this gives the condition \[ \boxed{a < \mu_s g \qquad \text{and} \quad a \leq \frac{b}{h} g} \qquad{[4]} \]
Step 4: Now we assume \(a > \frac{b}{h} g\) and the crate is assumed to tip about \(A\) in the counterclockwise direction. At time \(t=0^+\), the kinematics of the crate is characterized by \(\boldsymbol{a}_A = a\boldsymbol{\hat{\imath}}\) (\(A\) does not slip), and \(\boldsymbol{\alpha}= \alpha_0 \boldsymbol{\hat{k}}\) (we expect \(\alpha_0 >0\)). This gives the acceleration of mass center \(G\): \[ \boldsymbol{a}_G= \boldsymbol{a}_A + \alpha_0 \boldsymbol{\hat{k}}\times \boldsymbol{r}_{AG}= ( a -\frac{h}{2} \alpha_0 )\boldsymbol{\hat{\imath}}+ \frac{b}{2} \alpha_0 \boldsymbol{\hat{\jmath}} \] With the FBD diagram shown with the contact force acting through \(A\) (let \(x=b/2\)), the equations [1-3] now become: \[ m (a -\frac{h}{2} \alpha_0 ) = F, \qquad m \frac{b}{2} \alpha_0 = N-mg ,\qquad m \frac{b^2+h^2}{12} \alpha_0 = \frac{h}{2} F - \frac{b}{2} N \qquad{[5-7]} \] This easily gives \[ \boxed{ \alpha_0 = \frac{3}{2(h^2+b^2)} (h a - b g) }\qquad{[8]} \] This is valid as long as the static friction coefficient is sufficiently large: \(F <\mu_s N\). Using the expressions of \(F\) and \(N\) found in [5-7], we would find a condition \(\mu_s > \mu_{min}\), where an expression of \(\mu_{min}\) is a function of \((b,h, a, g)\).
If \(\mu <\mu_{min}\) and \(a > \frac{b}{h} g\), the crate would tip and slip.
Conclusion:
A. is correct: this is the condition stated in equation [4].
B. is incorrect: the correct condition is \(a < g \frac{b}{h}\).
C. is correct: this is the result found in stated in [8].
D. is incorrect: this would be obtained by setting \(\boldsymbol{a}_G = a\boldsymbol{\hat{\imath}}\).
E. is incorrect: the crate can simultaneously tip and slip.
Problem 2
A spherical body of radius \(r\), of uniform mass \(m\) and mass center \(G\), is given an initial spin at angular velocity \(\omega_0\) in the clockwise direction. It is then released on a straight support which is inclined at angle \(\theta\) with the horizontal. Mass center \(G\) has zero initial velocity.
The body is then observed to roll along the incline upward (in case \(\theta > 0\) shown in the figure below) or downward (in case \(\theta < 0\)).
Note that the moment of inertia of the body about one of its diameters is \(\frac{2}{5}mr^2\). The coefficient of (static or kinetic) friction is denoted as \(\mu\).
Answers
A. The body will always slip initially.
B. The body stops slipping at time \(t_1 =\frac{2 r\omega_0}{7\mu g\cos\theta}\).
C. The body will never stop slipping if the support is inclined downward \((\theta < 0)\).
D. The angular momentum \({\bf H}_I\) of the body about contact point \(I\) is conserved.
E. After rolling upward (for \(\theta > 0\)), the body will eventually stop its upward motion, and roll downward without slipping.
Solution
To figure out which given answers are correct, the problem requires a complete solution.
Step 1: Define a basis \((\boldsymbol{\hat{\imath}},\boldsymbol{\hat{\jmath}},\boldsymbol{\hat{k}})\) as shown in the figure. Define \(\boldsymbol{v}_G = v(t) \boldsymbol{\hat{\imath}}\) and \(\boldsymbol{\omega}= -\omega(t) \boldsymbol{\hat{k}}\) and the no-slip velocity \[ \boldsymbol{v}_{I\in 1/0}= v \boldsymbol{\hat{\imath}}-\omega\boldsymbol{\hat{k}}\times \boldsymbol{r}_{GI} = (v - r\omega) \boldsymbol{\hat{\imath}} \] At \(t=0\), \(v =0\) and \(\omega=\omega_0\). So the initial slip velocity is non-zero and points in the \((-\boldsymbol{\hat{\imath}})\)-direction, most likely until slipping stops at some \(t=t_1\).
Step 2: Sketch a FBD. Make sure that the friction force points in the \((+\boldsymbol{\hat{\imath}})\)-direction to oppose the slip velocity.
Step 3: To find time \(t=t_1\), we apply Euler’s principles \(m\boldsymbol{a}_G= F\boldsymbol{\hat{\imath}}+ N\boldsymbol{\hat{\jmath}}+ m{\bf g}\) and \(d{\bf H}_G/dt = {\bf M}_G\) to find 3 equations \[ m \dot{v}= F -mg \sin\theta, \quad N=mg\cos\theta,\quad -\frac{2}{5} mr^2 \dot{\omega}= r F \qquad{[1-3]} \] with \(F=\mu N= \mu m g\cos\theta\). We can integrate to find \(v\) and \(\omega\) vs \(t\): \[ \boxed{v = g( \mu \cos\theta-\sin\theta) t, \qquad r \omega= r\omega_0 - \frac{5}{2} \mu g t \cos\theta} \] At time \(t=t_1\) we set \(v - r\omega=0\): this gives \(t_1\) \[ \boxed{t_1 = \frac{r \omega_0}{ g( \frac{7}{2} \mu \cos\theta-\sin\theta)}} \] This result is true whether \(\theta>0\) or \(\theta<0\). In both cases, the body eventually stops slipping. Since the body is observed to move forward, this implies that \(\mu\) is sufficiently large \((\mu > \tan\theta)\): in other words \(t_1\) is not negative.
Step 4. For \(t\geq t_1\), all equations [1-3] remain the same, except that \(F=\mu N\) is now replaced by \(v= r\omega\) (no slip). Now we find \[ r\dot{\omega}= \dot{v}= - \frac{5}{7} g \sin\theta \] which gives (denoting \(v_1 = v(t_1)\)) \[ \boxed{v = v_1 - \frac{5}{7} g (t-t_1) \sin\theta} \] At some time \(t=t_2 > t_1\), \(v=0\) and the body eventually stops its upward motion (if \(\theta>0\)) and rolls downward without slipping (we recover that \(t_2 \to \infty\) when \(\theta=0\)).
Conclusion:
A. is correct, as shown above: the body slips initially (and eventually stops slipping at some time \(t_1\)).
B. is not quite right The answer for \(t_1\) would be correct for \(\theta=0\).
C. is incorrect: we found that \(t_1\) exists even for \(\theta<0\) (downward motion).
E. is correct: we found that for some time \(t_2 >t_1\), \(v=0\) (in case \(\theta>0\)): the body eventually stops its upward motion and rolls downward without slipping.
What about D? To answer D, we would have to apply \(d{\bf H}_I /dt +\boldsymbol{v}_I \times m \boldsymbol{v}_G = {\bf M}_I\). However \({\bf M}_I \neq 0\) (due to gravity). D is incorrect.
Problem 3
A pencil of mass \(m\) and length \(l\) is placed in the vertical position with its eraser in contact with the support at a point \(O\). It is then released without initial speed.
Assuming that the eraser remains fixed at \(O\) until the pencil reaches the horizontal position, and treating the pencil as a uniform slender rod (of length \(l\), mass \(m\), mass center \(G\), and mass moment of inertia \(I_{Gzz} = \frac{1}{12} ml^2)\),
Answers
A. The magnitude of the acceleration of extremity \(A\) becomes larger than the gravitational \(g\) at some point during the motion.
B. The maximum angular velocity reached by the pencil is \(\omega= \sqrt{3g/4l}\).
C. The reaction force at \(O\) vanishes at angle \(\theta_1 = \cos^{-1} (2/3)\) (measured relative to the vertical).
D. These answers would be independent of the actual mass distribution of the pencil (of length \(l\)).
Solution
To figure out which given answers are correct, the problem requires a complete solution.
Step 1: We adopt a basis \((\boldsymbol{\hat{\imath}}, \boldsymbol{\hat{\jmath}},\boldsymbol{\hat{k}})\) of fixed unit vectors. The motion of the body is rotational about point \(O\). Its position and kinematics is entirely defined by angle \(\theta\) as defined by the figure. At \(t=0\), \(\theta=0\) and \(\dot{\theta}=0\). We can also define a basis \((\boldsymbol{\hat{u}}, \boldsymbol{\hat{v}})\) attached to the body. The kinematics of the body is given by \(\boldsymbol{\omega}= -\dot{\theta}\boldsymbol{\hat{k}}\) and \(\boldsymbol{v}_G = \frac{l}{2} \dot{\theta}\boldsymbol{\hat{v}}\). This gives the acceleration \[ \boldsymbol{a}_G = \frac{l}{2} \ddot{\theta}\boldsymbol{\hat{v}}- \frac{l}{2} \dot{\theta}^2 \boldsymbol{\hat{u}} \]
Step 2: We sketch a FBD as shown in the figure.
Step 3: We first apply Euler’s second principle. Since the reaction of the support on the pencil has no moment about \(O\), we choose point \(O\) to evaluate moments and angular momentum \[ {\bf M}_O = \frac{d}{dt}{\bf H}_O \] with moment \({\bf M}_O = -mg \frac{l}{2} \sin\theta \boldsymbol{\hat{k}}\) and angular momentum \({\bf H}_O = -I_{Ozz} \dot{\theta} \boldsymbol{\hat{k}}\). With \(I_{Ozz} = I_{Gzz}+ ml^2/4= ml^2/3\), this gives the angular acceleration of the rod: \[ \ddot{\theta} = \frac{3g}{2l}\sin\theta \] and by integration (first pre-multiply by \(\dot{\theta}\)) \[ \boxed{ \dot{\theta}^2 = \frac{3g}{l}(1-\cos\theta) } \] using the initial condition \(\dot{\theta}=0\) at \(\theta =0\). These quantities take their maximum value at \(\theta=\pi/2\) (when the rod is horizontal, as expected): \[ \dot{\theta}_{max}^2 =\frac{3g}{l} , \qquad \ddot{\theta}_{max} = \frac{3g}{2l} \] This gives the acceleration of endpoint \(A\) at \(\theta=\pi/2\) \[ {\bf a}_A = -l\ddot{\theta} \boldsymbol{\hat{\jmath}}- l\dot{\theta}^2 \boldsymbol{\hat{\imath}}= -3g (\frac{1}{2}\boldsymbol{\hat{\jmath}}+\boldsymbol{\hat{\imath}}) \] with magnitude \[ \boxed{ |{\bf a}_A|_{\theta= \pi/2} = \frac{3\sqrt{5}}{2} g > g } \]
Step 4: We need to find the reaction at \(O\). We apply Euler’s first principle: \[ {\bf R}_O = R_x \boldsymbol{\hat{\imath}}+ R_y \boldsymbol{\hat{\jmath}}= m \boldsymbol{a}_G + mg \boldsymbol{\hat{\jmath}}= \frac{1}{2}ml \ddot{\theta}\boldsymbol{\hat{v}}- \frac{1}{2}ml \dot{\theta}^2 \boldsymbol{\hat{u}}+ mg \boldsymbol{\hat{\jmath}} \] We can use the expressions of \(\ddot{\theta}\) and \(\dot{\theta}^2\) found above in Step 3: \[ {\bf R}_O = \frac{3}{4} mg \sin\theta\boldsymbol{\hat{v}}- \frac{3}{2} mg (1-\cos\theta) \boldsymbol{\hat{u}}+ mg (\cos\theta\boldsymbol{\hat{u}}-\sin\theta\boldsymbol{\hat{v}}) \] This simplifies to give \[ {\bf R}_O = - \frac{1}{4} mg \sin\theta\boldsymbol{\hat{v}}+\frac{1}{2} mg (-3 + 5 \cos\theta) \boldsymbol{\hat{u}} \] Since the components do not both vanish at the same angle, we see that the reaction force never vanishes.
Note: We may express \({\bf R}_O\) on \((\boldsymbol{\hat{\imath}},\boldsymbol{\hat{\jmath}})\) to find \[ {\bf R}_O = - \frac{3}{4} mg (3 \cos\theta-2) \boldsymbol{\hat{\imath}}+\frac{1}{4} mg (1- 3 \cos\theta)^2 \boldsymbol{\hat{\jmath}} \]
Conclusion:
A. is correct: We found \(|a_A|_{max} > g\) at the value \(\theta=\pi/2\).
B. is incorrect: We found \(\omega_{max} =\sqrt{3g/l}\).
C. is incorrect: The reaction force at \(O\) never vanishes.
D. is incorrect: These results depend on the mass distribution of the slender rod. The mass distribution affects the location of \(G\) and the value of moment \(I_{Ozz}\).