Solution to Question 19
1. Kinematics of Separation
We define the table edge at $X=0$. Initially, the glass is at $X=x$, and the cloth extends from $X=0$ to $X=l$. The cloth is pulled to the left (off the table) with speed $v_c$.
Glass Motion: Kinetic friction $\mu mg$ acts on the glass in the direction of the cloth’s motion (left).
$$ a_g = \mu g $$
Position of glass: $X_g(t) = x – \frac{1}{2} \mu g t^2$.
Cloth Motion: The right end of the cloth (originally at $l$) moves left.
$$ X_{cloth}(t) = l – v_c t $$
Separation Condition: The glass leaves the cloth when the right end of the cloth passes underneath it ($X_g = X_{cloth}$).
$$ x – \frac{1}{2} \mu g t^2 = l – v_c t $$
$$ \frac{1}{2} \mu g t^2 – v_c t + (l-x) = 0 $$
2. Conditions for Success
For a successful demonstration, the glass must not fall off the table. This means at the moment of separation $t$, the glass must still be on the table ($X_g(t) > 0$).
$$ x – \frac{1}{2} \mu g t^2 > 0 \implies x > \frac{1}{2} \mu g t^2 $$
Substituting $(l-x) = v_c t – \frac{1}{2} \mu g t^2$ into the condition:
$$ x > v_c t – (l-x) \implies v_c t < l \implies t < \frac{l}{v_c} $$
We solve the quadratic equation for the separation time $t$:
$$ t = \frac{v_c - \sqrt{v_c^2 - 2\mu g(l-x)}}{\mu g} $$
We require two things:
- A real solution exists (Glass actually separates): $v_c^2 \ge 2\mu g(l-x)$.
- The safety condition holds: $t < l/v_c$.
3. Deriving the Velocity Range
Substitute $t$ into the inequality:
$$ \frac{v_c – \sqrt{v_c^2 – 2\mu g(l-x)}}{\mu g} < \frac{l}{v_c} $$
$$ v_c^2 - v_c \sqrt{v_c^2 - 2\mu g(l-x)} < \mu g l $$
$$ v_c^2 - \mu g l < v_c \sqrt{v_c^2 - 2\mu g(l-x)} $$
Squaring both sides (valid if LHS > 0, i.e., $v_c^2 > \mu g l$):
$$ (v_c^2 – \mu g l)^2 < v_c^2 (v_c^2 - 2\mu g(l-x)) $$
$$ v_c^4 - 2v_c^2 \mu g l + \mu^2 g^2 l^2 < v_c^4 - 2v_c^2 \mu g l + 2 v_c^2 \mu g x $$
$$ \mu^2 g^2 l^2 < 2 v_c^2 \mu g x $$
$$ v_c^2 > \frac{\mu g l^2}{2x} \implies v_c > \sqrt{\frac{\mu g l^2}{2x}} $$
Case 1 ($x < l/2$): The condition $v_c > \sqrt{\frac{\mu g l^2}{2x}}$ implies $v_c^2 > \mu g l$ is automatically satisfied. So the lower bound is $\sqrt{\frac{\mu g l^2}{2x}}$.
Case 2 ($x > l/2$): Here, the constraint comes from the existence of the solution (discriminant $\ge 0$): $v_c^2 \ge 2\mu g(l-x)$. The safety condition is automatically satisfied for all real $t$ in this range.
Answer:
$$ v_c > \sqrt{\frac{\mu g l^2}{2x}} \quad \text{for } x < \frac{l}{2} $$
$$ v_c > \sqrt{2\mu g(l-x)} \quad \text{for } x > \frac{l}{2} $$
We define the table edge at $X=0$. Initially, the glass is at $X=x$, and the cloth extends from $X=0$ to $X=l$. The cloth is pulled to the left (off the table) with speed $v_c$.
Glass Motion: Kinetic friction $\mu mg$ acts on the glass in the direction of the cloth’s motion (left). $$ a_g = \mu g $$ Position of glass: $X_g(t) = x – \frac{1}{2} \mu g t^2$.
Cloth Motion: The right end of the cloth (originally at $l$) moves left. $$ X_{cloth}(t) = l – v_c t $$
Separation Condition: The glass leaves the cloth when the right end of the cloth passes underneath it ($X_g = X_{cloth}$). $$ x – \frac{1}{2} \mu g t^2 = l – v_c t $$ $$ \frac{1}{2} \mu g t^2 – v_c t + (l-x) = 0 $$
For a successful demonstration, the glass must not fall off the table. This means at the moment of separation $t$, the glass must still be on the table ($X_g(t) > 0$). $$ x – \frac{1}{2} \mu g t^2 > 0 \implies x > \frac{1}{2} \mu g t^2 $$ Substituting $(l-x) = v_c t – \frac{1}{2} \mu g t^2$ into the condition: $$ x > v_c t – (l-x) \implies v_c t < l \implies t < \frac{l}{v_c} $$ We solve the quadratic equation for the separation time $t$: $$ t = \frac{v_c - \sqrt{v_c^2 - 2\mu g(l-x)}}{\mu g} $$ We require two things:
- A real solution exists (Glass actually separates): $v_c^2 \ge 2\mu g(l-x)$.
- The safety condition holds: $t < l/v_c$.
Substitute $t$ into the inequality:
$$ \frac{v_c – \sqrt{v_c^2 – 2\mu g(l-x)}}{\mu g} < \frac{l}{v_c} $$
$$ v_c^2 - v_c \sqrt{v_c^2 - 2\mu g(l-x)} < \mu g l $$
$$ v_c^2 - \mu g l < v_c \sqrt{v_c^2 - 2\mu g(l-x)} $$
Squaring both sides (valid if LHS > 0, i.e., $v_c^2 > \mu g l$):
$$ (v_c^2 – \mu g l)^2 < v_c^2 (v_c^2 - 2\mu g(l-x)) $$
$$ v_c^4 - 2v_c^2 \mu g l + \mu^2 g^2 l^2 < v_c^4 - 2v_c^2 \mu g l + 2 v_c^2 \mu g x $$
$$ \mu^2 g^2 l^2 < 2 v_c^2 \mu g x $$
$$ v_c^2 > \frac{\mu g l^2}{2x} \implies v_c > \sqrt{\frac{\mu g l^2}{2x}} $$
Case 1 ($x < l/2$): The condition $v_c > \sqrt{\frac{\mu g l^2}{2x}}$ implies $v_c^2 > \mu g l$ is automatically satisfied. So the lower bound is $\sqrt{\frac{\mu g l^2}{2x}}$.
Case 2 ($x > l/2$): Here, the constraint comes from the existence of the solution (discriminant $\ge 0$): $v_c^2 \ge 2\mu g(l-x)$. The safety condition is automatically satisfied for all real $t$ in this range.