Solution to Question 33
Let the inclined plane be the $x-y$ plane.
• $y$-axis: Along the line of fastest descent (down the slope).
• $x$-axis: Horizontal line on the plane.
Let $\vec{v}$ be the instantaneous velocity making an angle $\phi$ with the $y$-axis.
To solve this, we must connect the change in horizontal velocity $v_x$ to the change in angle $\phi$.
Step A: Forces and Accelerations
The forces acting in the plane are Gravity ($mg\sin\theta$ along $y$) and Friction ($-\mu mg \cos\theta$ opposing $\vec{v}$).
Step B: Normal Acceleration
We decompose acceleration perpendicular to the velocity vector (normal component $a_n$). Gravity has a component $g \sin\theta \sin\phi$ perpendicular to $\vec{v}$. Friction is antiparallel to velocity, so it has no normal component.
Step C: Combining Equations
Now we substitute $dt$ from (2) into equation (1) for $dv_x$:
$$ dv_x = a_x dt = (-\mu g \cos\theta \sin\phi) \left( \frac{-v \, d\phi}{g \sin\theta \sin\phi} \right) $$ $$ dv_x = \frac{\mu \cos\theta}{\sin\theta} v \, d\phi = \mu \cot\theta \, v \, d\phi $$Using the kinematic relation $v_x = v \sin\phi \implies v = \frac{v_x}{\sin\phi}$:
$$ dv_x = \mu \cot\theta \left( \frac{v_x}{\sin\phi} \right) d\phi $$Let $k = \mu \cot\theta$. Integrating both sides:
$$ \ln v_x = k \ln(\tan(\phi/2)) + \text{constant} $$ $$ v_x = A \left[ \tan\left(\frac{\phi}{2}\right) \right]^k \quad \text{(3)} $$Here $A$ is a constant determined by initial conditions.
We need the total time $T$ for the block to stop. We integrate $dt$ using the expression for $a_x$.
$$ T = \int dt = \int_{v_{0x}}^{0} \frac{dt}{dv_x} dv_x = \int_{v_{0x}}^{0} \frac{1}{a_x} dv_x $$ $$ T = \int_{v_{0x}}^{0} \frac{dv_x}{-\mu g \cos\theta \sin\phi} = \frac{1}{\mu g \cos\theta} \int_{0}^{v_{0x}} \frac{dv_x}{\sin\phi} $$To perform this integration, we express $1/\sin\phi$ in terms of $v_x$ using the half-angle identity $\sin\phi = \frac{2\tan(\phi/2)}{1+\tan^2(\phi/2)}$. From eq (3), $\tan(\phi/2) = (v_x/A)^{1/k}$.
$$ \frac{1}{\sin\phi} = \frac{1 + (v_x/A)^{2/k}}{2(v_x/A)^{1/k}} = \frac{1}{2} \left[ \left(\frac{v_x}{A}\right)^{-1/k} + \left(\frac{v_x}{A}\right)^{1/k} \right] $$Note: The exponent $1/k = \frac{\tan\theta}{\mu}$. Let us define $\lambda = 1/k$.
$$ T = \frac{1}{2\mu g \cos\theta} \int_{0}^{v_{0x}} \left[ \left(\frac{A}{v_x}\right)^{\lambda} + \left(\frac{v_x}{A}\right)^{\lambda} \right] dv_x $$Integrating term by term:
$$ T = \frac{1}{2\mu g \cos\theta} \left[ A^\lambda \frac{v_x^{1-\lambda}}{1-\lambda} + A^{-\lambda} \frac{v_x^{1+\lambda}}{1+\lambda} \right]_{0}^{v_{0x}} $$Substitute the upper limit $v_{0x}$:
$$ T = \frac{v_{0x}}{2\mu g \cos\theta} \left[ \frac{(A/v_{0x})^\lambda}{1-\lambda} + \frac{(v_{0x}/A)^\lambda}{1+\lambda} \right] $$From eq (3), at the initial moment: $(v_{0x}/A)^\lambda = \tan(\phi_0/2)$ and $(A/v_{0x})^\lambda = \cot(\phi_0/2)$.
$$ T = \frac{v_{0x}}{2\mu g \cos\theta} \left[ \frac{\cot(\phi_0/2)}{1-\lambda} + \frac{\tan(\phi_0/2)}{1+\lambda} \right] $$We combine the terms inside the bracket. Let $t = \tan(\phi_0/2)$:
$$ \text{Bracket} = \frac{1/t}{1-\lambda} + \frac{t}{1+\lambda} = \frac{1+\lambda + t^2(1-\lambda)}{t(1-\lambda^2)} = \frac{(1+t^2) + \lambda(1-t^2)}{t(1-\lambda^2)} $$Recall $\sin\phi_0 = \frac{2t}{1+t^2}$ and $\cos\phi_0 = \frac{1-t^2}{1+t^2}$. Multiplying numerator and denominator by $\frac{1}{1+t^2}$:
$$ \text{Bracket} = \frac{1 + \lambda \cos\phi_0}{\frac{1}{2}\sin\phi_0 (1-\lambda^2)} = \frac{2(1 + \lambda \cos\phi_0)}{\sin\phi_0 (1-\lambda^2)} $$Substitute this back into the expression for $T$ (using $v_{0x} = v_0 \sin\phi_0$):
$$ T = \frac{v_0 \sin\phi_0}{2\mu g \cos\theta} \cdot \frac{2(1 + \lambda \cos\phi_0)}{\sin\phi_0 (1-\lambda^2)} = \frac{v_0 (1 + \lambda \cos\phi_0)}{\mu g \cos\theta (1-\lambda^2)} $$Finally, substitute $\lambda = \frac{\tan\theta}{\mu} = \frac{\sin\theta}{\mu \cos\theta}$:
- Denominator factor: $1-\lambda^2 = \frac{\mu^2\cos^2\theta – \sin^2\theta}{\mu^2\cos^2\theta}$
- Numerator factor: $1+\lambda\cos\phi_0 = \frac{\mu\cos\theta + \sin\theta\cos\phi_0}{\mu\cos\theta}$
Plugging these in simplifies to:
$$ T = \frac{v_0 (\mu \cos\theta + \sin\theta \cos\phi_0)}{g(\mu^2 \cos^2\theta – \sin^2\theta)} $$