No, that's not how that works. If tip is ground and ring is Vcc, there is no possible way to create a dangerous short. So lets imagine what happens to the host

No, that's not how that works. If tip is ground and ring is Vcc, there is no possible way to create a dangerous short. So lets imagine what happens to the host port when you unplug a live device. Vcc breaks first, and all other pins shift. D+ and D- are safe to short together or to ground (and I'm pretty sure safe to short to Vcc) so they don't matter much. The shield is physically out of the socket. As the connectors move, you get connections that put Vcc on the device's Data pins, through the device ground to host Data pins. This is equivalent to a large resistor between host Vcc and data pins, which is safe. As the connector is removed, the device's ground contacts host's Vcc. Nothing happens because all the other pins are physically outside of the socket. There is no circuit, no path for the current to go. The only way it could be a problem is if the device had some external ground connection. That would create a direct short between the host's Vcc and ground. But (correctly implemented) USB ports can safely tolerate a short from Vcc to ground. If not, it's not complicated or expensive to build protection into the device. Even if all of the above weren't true, it's still not complicated or expensive to design a port which is tolerant to any kind of short on any pins. What you're arguing is both not possible and wouldn't be a problem even if it were.